Ch-4 Exploring Algebraic Identities class 9th maths ganita manjari ncert solution

Exercise Set 4.1

(i) $(7x+4y)^2$

Here $a=7x,\;b=4y$
$(7x+4y)^2=(7x)^2+2(7x)(4y)+(4y)^2=49x^2+56xy+16y^2$

Answer$[\ 49x^2+56xy+16y^2\ ]$

(ii) $\left(\dfrac{7}{5}x+\dfrac{3}{2}y\right)^2$

$a=\dfrac{7}{5}x,\;b=\dfrac{3}{2}y$

$=\dfrac{49}{25}x^2+2\cdot\dfrac{7}{5}x\cdot\dfrac{3}{2}y+\dfrac{9}{4}y^2=\dfrac{49}{25}x^2+\dfrac{21}{5}xy+\dfrac{9}{4}y^2$

Answer$\left[\ \dfrac{49}{25}x^2+\dfrac{21}{5}xy+\dfrac{9}{4}y^2\ \right]$

(iii) $(2.5p+1.5q)^2$

$=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2=6.25p^2+7.5pq+2.25q^2$

Answer$[\ 6.25p^2+7.5pq+2.25q^2\ ]$

(iv) $\left(\dfrac{3}{4}s+8t\right)^2$

$=\dfrac{9}{16}s^2+2\cdot\dfrac{3}{4}s\cdot8t+64t^2=\dfrac{9}{16}s^2+12st+64t^2$

Answer$\left[\ \dfrac{9}{16}s^2+12st+64t^2\ \right]$

(v) $\left(x+\dfrac{1}{2y}\right)^2$

$=x^2+2\cdot x\cdot\dfrac{1}{2y}+\dfrac{1}{4y^2}=x^2+\dfrac{x}{y}+\dfrac{1}{4y^2}$

Answer$\left[\ x^2+\dfrac{x}{y}+\dfrac{1}{4y^2}\ \right]$

(vi) $\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2$

$=\dfrac{1}{x^2}+\dfrac{2}{xy}+\dfrac{1}{y^2}$

Answer$\left[\ \dfrac{1}{x^2}+\dfrac{2}{xy}+\dfrac{1}{y^2}\ \right]$

(i) $64^2$

$64=60+4$
$(60+4)^2=3600+480+16=4096$

Answer$[\ 4096\ ]$

(ii) $105^2$

$105=100+5$
$(100+5)^2=10000+1000+25=11025$

Answer$[\ 11025\ ]$

(iii) $205^2$

$205=200+5$
$(200+5)^2=40000+2000+25=42025$

Answer$[\ 42025\ ]$

Exercise Set 4.2

(i) $9x^2+24xy+16y^2$

$(3x)^2+2(3x)(4y)+(4y)^2$; matches $(a+b)^2$ with $a=3x,b=4y$

Answer$[\ (3x+4y)^2\ ]$

(ii) $4s^2+20st+25t^2$

$(2s)^2+2(2s)(5t)+(5t)^2$; $a=2s,b=5t$

Answer$[\ (2s+5t)^2\ ]$

(iii) $49x^2+28xy+4y^2$

$(7x)^2+2(7x)(2y)+(2y)^2$; $a=7x,b=2y$

Answer$[\ (7x+2y)^2\ ]$

(iv) $64p^2+\dfrac{32}{3}pq+\dfrac{4}{9}q^2$

$(8p)^2+2(8p)\!\left(\dfrac{2}{3}q\right)+\left(\dfrac{2}{3}q\right)^2$; $a=8p,\;b=\dfrac{2}{3}q$

Answer$\left[\ \left(8p+\dfrac{2}{3}q\right)^2\ \right]$

(v)★ $3a^2+4ab+\dfrac{4}{3}b^2$

Factor out 3: $3\!\left(a^2+\dfrac{4}{3}ab+\dfrac{4}{9}b^2\right)=3\!\left(a+\dfrac{2}{3}b\right)^2$

Answer$\left[\ 3\!\left(a+\dfrac{2}{3}b\right)^2\ \right]$

(vi)★ $\dfrac{9}{5}s^2+6sv+5v^2$

Factor out $\dfrac{1}{5}$: $\dfrac{1}{5}(9s^2+30sv+25v^2)=\dfrac{1}{5}(3s+5v)^2$

Answer$\left[\ \dfrac{1}{5}(3s+5v)^2\ \right]$

(i) $79^2$

$79=80-1$
$(80-1)^2=6400-160+1=6241$

Answer$[\ 6241\ ]$

(ii) $193^2$

$193=200-7$
$(200-7)^2=40000-2800+49=37249$

Answer$[\ 37249\ ]$

(iii) $299^2$

$299=300-1$
$(300-1)^2=90000-600+1=89401$

Answer$[\ 89401\ ]$

Exercise Set 4.3

(i) $117^2$

$117=100+17$; use $(a+b)^2$
$10000+3400+289=13689$

Answer$[\ 13689\ ]$

(ii) $78^2$

$78=80-2$; use $(a-b)^2$
$6400-320+4=6084$

Answer$[\ 6084\ ]$

(iii) $198^2$

$198=200-2$
$40000-800+4=39204$

Answer$[\ 39204\ ]$

(iv) $214^2$

$214=200+14$
$40000+5600+196=45796$

Answer$[\ 45796\ ]$

(v) $1104^2$

$1104=1000+100+4$; use $(a+b+c)^2$
$1000000+10000+16+200000+800+8000=1218816$

Answer$[\ 1218816\ ]$

(vi) $1120^2$

$1120=1000+100+20$
$1000000+10000+400+200000+4000+40000=1254400$

Answer$[\ 1254400\ ]$

(i) $16y^2-24y+9$

$(4y)^2-2(4y)(3)+3^2$; $a=4y,b=3$

Answer$[\ (4y-3)^2\ ]$

(ii) $\dfrac{9}{4}s^2+6st+4t^2$

$\left(\dfrac{3}{2}s\right)^2+2\cdot\dfrac{3}{2}s\cdot2t+(2t)^2$; $a=\dfrac{3}{2}s,b=2t$

Answer$\left[\ \left(\dfrac{3}{2}s+2t\right)^2\ \right]$

(iii) $\dfrac{m^2}{9}+\dfrac{mk}{3}+\dfrac{k^2}{4}+3nk+2mn+9n^2$

Identify $a=\dfrac{m}{3},\;b=\dfrac{k}{2},\;c=3n$; matches $(a+b+c)^2$

Answer$\left[\ \left(\dfrac{m}{3}+\dfrac{k}{2}+3n\right)^2\ \right]$

(iv) $\dfrac{p^2}{16}-2+\dfrac{16}{p^2}$

$\left(\dfrac{p}{4}\right)^2-2\cdot\dfrac{p}{4}\cdot\dfrac{4}{p}+\left(\dfrac{4}{p}\right)^2$; $a=\dfrac{p}{4},b=\dfrac{4}{p}$

Answer$\left[\ \left(\dfrac{p}{4}-\dfrac{4}{p}\right)^2\ \right]$

(v) $9a^2+4b^2+c^2-12ab+6ac-4bc$

$a’=3a,\;b’=-2b,\;c’=c$; cross-terms: $-12ab$ ✓, $-4bc$ ✓, $6ac$ ✓

Answer$[\ (3a-2b+c)^2\ ]$

(i) $(p+3q+7r)^2$

$a=p,b=3q,c=7r$
$=p^2+9q^2+49r^2+6pq+42qr+14pr$

Answer$[\ p^2+9q^2+49r^2+6pq+42qr+14pr\ ]$

(ii) $(3x-2y+4z)^2$

$a=3x,b=-2y,c=4z$
$=9x^2+4y^2+16z^2-12xy-16yz+24xz$

Answer$[\ 9x^2+4y^2+16z^2-12xy-16yz+24xz\ ]$

Expand each square:
$(a+b-c)^2=a^2+b^2+c^2+2ab-2bc-2ca$
$(a-b+c)^2=a^2+b^2+c^2-2ab-2bc+2ca$
$(a-b-c)^2=a^2+b^2+c^2-2ab+2bc-2ca$

Adding LHS $=3a^2+3b^2+3c^2-2ab-2bc-2ca$

RHS $=2a^2+2b^2+2c^2$

LHS $\neq$ RHS. Counter-example: $a=b=c=1$ gives LHS $=3$, RHS $=6$.

Answer$[\ \text{NOT an identity. LHS} = 3a^2+3b^2+3c^2-2ab-2bc-2ca \neq \text{RHS}\ ]$

Exercise Set 4.4

(i) $s^2-11s+24=(\ \_\_\_\ )(\ \_\_\_\ )$

Need $a+b=-11,\;ab=24$ → $a=-8,b=-3$
$(s-8)(s-3)$

Answer$[\ (s-8)(s-3)\ ]$

(ii) $(\ \_\_\_\ )(x+1)=3x^2-4x-7$

Factor: $3x^2+3x-7x-7=3x(x+1)-7(x+1)=(3x-7)(x+1)$

Answer$[\ (3x-7)(x+1)\ ]$

(iii) $10x^2-11x-6=(2x-\ \_\_\_)(\ \_\_\_+2)$

$ac=-60$; choose $4$ and $-15$: $10x^2+4x-15x-6=2x(5x+2)-3(5x+2)=(2x-3)(5x+2)$
Blank 1 = $3$, Blank 2 = $5x$

Answer$[\ (2x-3)(5x+2)\ ]$

(iv) $6x^2+7x+2=(\ \_\_\_)(\ \_\_\_)$

$ac=12$; choose $3,4$: $6x^2+3x+4x+2=3x(2x+1)+2(2x+1)=(3x+2)(2x+1)$

Answer$[\ (3x+2)(2x+1)\ ]$

(i) $41^2$

$(40+1)^2=1600+80+1=1681$

Answer$[\ 1681\ ]$

(ii) $27^2$

$(20+7)^2=400+280+49=729$

Answer$[\ 729\ ]$

(iii) $23\times17$

$(20+3)(20-3)=400-9=391$

Answer$[\ 391\ ]$

(iv) $135^2$

$(130+5)^2=16900+1300+25=18225$

Answer$[\ 18225\ ]$

(v) $97^2$

$(100-3)^2=10000-600+9=9409$

Answer$[\ 9409\ ]$

(vi) $18\times29$

$(x+a)(x+b)$ with $x=20,a=-2,b=9$: $400+140-18=522$

Answer$[\ 522\ ]$

(vii) $34\times43$

$x=40,a=-6,b=3$: $1600-120-18=1462$

Answer$[\ 1462\ ]$

(viii) $205^2$

$(200+5)^2=40000+2000+25=42025$

Answer$[\ 42025\ ]$

(i) $9a^2+b^2+4c^2-6ab+12ac-4bc$

$a’=3a,b’=-b,c’=2c$; cross-terms $-6ab$ ✓, $-4bc$ ✓, $12ac$ ✓

Answer$[\ (3a-b+2c)^2\ ]$

(ii) $16s^2+25t^2-40st$

$(4s)^2-2(4s)(5t)+(5t)^2$; $a=4s,b=5t$

Answer$[\ (4s-5t)^2\ ]$

(iii) $r^2-r-42$

$a+b=-1,ab=-42$ → $a=6,b=-7$

Answer$[\ (r+6)(r-7)\ ]$

(iv) $49g^2+14gh+h^2$

$(7g)^2+2(7g)(h)+h^2$; $a=7g,b=h$

Answer$[\ (7g+h)^2\ ]$

(v) $64u^2+121v^2+4w^2-176uv-32uw+44vw$

$a’=8u,b’=-11v,c’=-2w$: $-176uv$ ✓, $44vw$ ✓, $-32uw$ ✓

Answer$[\ (8u-11v-2w)^2\ ]$

Exercise Set 4.5

(i) $\dfrac{3p^2-3pq-18q^2}{p^2+3pq-10q^2}$

Num: $3(p+2q)(p-3q)$; Den: $(p-2q)(p+5q)$
No common factor.

Answer$\left[\ \dfrac{3(p+2q)(p-3q)}{(p-2q)(p+5q)}\ \right]$

(ii) $\dfrac{n^3-3n^2m+3nm^2-m^3}{5m^2-10mn+5n^2}$

Num: $(n-m)^3$; Den: $5(n-m)^2$
$=\dfrac{n-m}{5}$

Answer$\left[\ \dfrac{n-m}{5}\ \right]$

(iii) $\dfrac{w^3-v^3+x^3+3wvx}{w^2+v^2+x^2-2wv-2vx+2wx}$

Num: $(w-v+x)(w^2+v^2+x^2+wv+vx-wx)$ (identity with $b=-v$)
Den: $(w-v+x)^2$
$=\dfrac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}$

Answer$\left[\ \dfrac{w^2+v^2+x^2+wv+vx-wx}{w-v+x}\ \right]$

(iv) $\dfrac{4y^2-20yz+25z^2}{25z^2-4y^2}$

Num: $(2y-5z)^2$; Den: $(5z-2y)(5z+2y)=-(2y-5z)(5z+2y)$
$=\dfrac{5z-2y}{5z+2y}$

Answer$\left[\ \dfrac{5z-2y}{5z+2y}\ \right]$

(v) $\dfrac{(x^2+x-6)(x^2-7x+12)}{(x^2-6x+8)(x^2-9)}$

Factor: $\dfrac{(x+3)(x-2)(x-3)(x-4)}{(x-2)(x-4)(x-3)(x+3)}$; all cancel $=1$

Answer$[\ 1\ ]$

(vi) $\dfrac{p^4-16}{p^2-4p+4}$

Num: $(p^2-4)(p^2+4)=(p-2)(p+2)(p^2+4)$
Den: $(p-2)^2$
$=\dfrac{(p+2)(p^2+4)}{p-2}$

Answer$\left[\ \dfrac{(p+2)(p^2+4)}{p-2}\ \right]$

End-of-Chapter Exercises

(i) $(-3x+4)^2$

$(4-3x)^2=(4)^2-2(4)(3x)+(3x)^2=16-24x+9x^2$

Answer$[\ 9x^2-24x+16\ ]$

(ii) $(2s+7)(2s-7)$

$(a+b)(a-b)$: $a=2s,b=7$; $=4s^2-49$

Answer$[\ 4s^2-49\ ]$

(iii) $\left(p^2+\tfrac{1}{2}\right)\!\left(p^2-\tfrac{1}{2}\right)$

$a=p^2,b=\tfrac{1}{2}$; $=p^4-\tfrac{1}{4}$

Answer$\left[\ p^4-\dfrac{1}{4}\ \right]$

(iv) $(2n+7)(2n-7)$

$=4n^2-49$

Answer$[\ 4n^2-49\ ]$

(v) $(s-2t)(s^2+2st+4t^2)$

$a^3-b^3$ identity: $a=s,b=2t$; $=s^3-8t^3$

Answer$[\ s^3-8t^3\ ]$

(vi) $\left(\tfrac{1}{2r}-4r\right)^2$

$=\dfrac{1}{4r^2}-2\cdot\dfrac{1}{2r}\cdot4r+16r^2=\dfrac{1}{4r^2}-4+16r^2$

Answer$\left[\ \dfrac{1}{4r^2}-4+16r^2\ \right]$

(vii) $(-3m+4k-l)^2$

$a=-3m,b=4k,c=-l$
$=9m^2+16k^2+l^2-24mk-8kl+6lm$

Answer$[\ 9m^2+16k^2+l^2-24mk-8kl+6lm\ ]$

(viii) $\left(x-\tfrac{1}{3}y\right)^3$

$(a-b)^3$: $a=x,b=\tfrac{y}{3}$
$=x^3-x^2y+\dfrac{xy^2}{3}-\dfrac{y^3}{27}$

Answer$\left[\ x^3-x^2y+\dfrac{xy^2}{3}-\dfrac{y^3}{27}\ \right]$

(ix) $\left(\tfrac{7}{2}k-\tfrac{2}{3}m\right)^3$

$a=\tfrac{7k}{2},b=\tfrac{2m}{3}$; $(a-b)^3$
$=\dfrac{343k^3}{8}-\dfrac{49k^2m}{2}+\dfrac{14km^2}{3}-\dfrac{8m^3}{27}$

Answer$\left[\ \dfrac{343k^3}{8}-\dfrac{49k^2m}{2}+\dfrac{14km^2}{3}-\dfrac{8m^3}{27}\ \right]$

(i) $17\times21$

$(19-2)(19+2)=361-4=357$

Answer$[\ 357\ ]$

(ii) $104\times96$

$(100+4)(100-4)=10000-16=9984$

Answer$[\ 9984\ ]$

(iii) $24\times16$

$(20+4)(20-4)=400-16=384$

Answer$[\ 384\ ]$

(iv) $147^3$

$(150-3)^3=3375000-202500+4050-27=3176523$

Answer$[\ 3176523\ ]$

(v) $199^3$

$(200-1)^3=8000000-120000+600-1=7880599$

Answer$[\ 7880599\ ]$

(vi) $127^3$

$(130-3)^3=2197000-152100+3510-27=2048383$

Answer$[\ 2048383\ ]$

(vii) $(-107)^3$

$-(107)^3=-(100+7)^3=-(1000000+210000+14700+343)=-1225043$

Answer$[\ -1225043\ ]$

(viii) $(-299)^3$

$-(300-1)^3=-(27000000-270000+900-1)=-26730899$

Answer$[\ -26730899\ ]$

(i) $4y^2+1+\dfrac{1}{16y^2}$

$(2y)^2+2(2y)\!\left(\dfrac{1}{4y}\right)+\left(\dfrac{1}{4y}\right)^2$

Answer$\left[\ \left(2y+\dfrac{1}{4y}\right)^2\ \right]$

(ii) $9m^2-\dfrac{1}{25n^2}$

$(3m)^2-\left(\dfrac{1}{5n}\right)^2$

Answer$\left[\ \left(3m-\dfrac{1}{5n}\right)\!\left(3m+\dfrac{1}{5n}\right)\ \right]$

(iii) $27b^3-\dfrac{1}{64b^3}$

$(3b)^3-\left(\dfrac{1}{4b}\right)^3$; use $a^3-b^3$

Answer$\left[\ \left(3b-\dfrac{1}{4b}\right)\!\left(9b^2+\dfrac{3}{4}+\dfrac{1}{16b^2}\right)\ \right]$

(iv) $x^2+\dfrac{5x}{6}+\dfrac{1}{6}$

$a+b=\dfrac{5}{6},\;ab=\dfrac{1}{6}$ → $a=\dfrac{1}{2},b=\dfrac{1}{3}$

Answer$\left[\ \left(x+\dfrac{1}{2}\right)\!\left(x+\dfrac{1}{3}\right)\ \right]$

(v) $27u^3-\dfrac{1}{125}-\dfrac{27u^2}{5}+\dfrac{9u}{25}$

$(3u-\tfrac{1}{5})^3$: check $27u^3$ ✓, $\dfrac{27u^2}{5}$ ✓, $\dfrac{9u}{25}$ ✓, $\dfrac{1}{125}$ ✓

Answer$\left[\ \left(3u-\dfrac{1}{5}\right)^3\ \right]$

(vi) $64y^3+\dfrac{z^3}{125}$

$(4y)^3+\left(\dfrac{z}{5}\right)^3$; use $a^3+b^3$

Answer$\left[\ \left(4y+\dfrac{z}{5}\right)\!\left(16y^2-\dfrac{4yz}{5}+\dfrac{z^2}{25}\right)\ \right]$

(vii) $p^3+27q^3+r^3-9pqr$

$p^3+(3q)^3+r^3-3p(3q)r$; identity with $a=p,b=3q,c=r$

Answer$[\ (p+3q+r)(p^2+9q^2+r^2-3pq-3qr-pr)\ ]$

(viii) $9m^2-12m+4$

$(3m-2)^2$

Answer$[\ (3m-2)^2\ ]$

(ix) $9x^3-\dfrac{8y^3}{3}+\dfrac{z^3}{3}+6xyz$

$=\dfrac{1}{3}(27x^3-8y^3+z^3+18xyz)=\dfrac{1}{3}[(3x)^3+(-2y)^3+z^3-3(3x)(-2y)(z)]$
Use identity with $a=3x,b=-2y,c=z$

Answer$\left[\ \dfrac{1}{3}(3x-2y+z)(9x^2+4y^2+z^2+6xy+2yz-3xz)\ \right]$

(x) $4x^2+9y^2+36z^2+12xy+36yz+24xz$

$a=2x,b=3y,c=6z$: $(2x+3y+6z)^2$

Answer$[\ (2x+3y+6z)^2\ ]$

(xi) $27u^3-\dfrac{1}{216}-\dfrac{9u^2}{2}+\dfrac{u}{4}$

$(3u-\tfrac{1}{6})^3$: $27u^3$ ✓, $\dfrac{9u^2}{2}$ ✓, $\dfrac{u}{4}$ ✓, $\dfrac{1}{216}$ ✓

Answer$\left[\ \left(3u-\dfrac{1}{6}\right)^3\ \right]$

(i) $\dfrac{4x^2+4x+1}{4x^2-1}$

Num: $(2x+1)^2$; Den: $(2x-1)(2x+1)$; cancel $(2x+1)$

Answer$\left[\ \dfrac{2x+1}{2x-1}\ \right]$

(ii) $\dfrac{9(3a^3-24b^3)}{9a^2-36b^2}$

Num: $27(a-2b)(a^2+2ab+4b^2)$; Den: $9(a-2b)(a+2b)$; cancel

Answer$\left[\ \dfrac{3(a^2+2ab+4b^2)}{a+2b}\ \right]$

(iii) $\dfrac{s^3+125t^3}{s^2-2st-35t^2}$

Num: $(s+5t)(s^2-5st+25t^2)$; Den: $(s+5t)(s-7t)$; cancel

Answer$\left[\ \dfrac{s^2-5st+25t^2}{s-7t}\ \right]$

(i) Area $=25a^2-30ab+9b^2$ sq. units

$(5a-3b)^2=(5a-3b)(5a-3b)$; it is a square

Answer$[\ \text{Length} = \text{Breadth} = (5a-3b)\text{ units}\ ]$

(ii) Area $=36s^2-49t^2$ sq. units

$(6s-7t)(6s+7t)$

Answer$[\ \text{Length}=(6s+7t),\;\text{Breadth}=(6s-7t)\text{ units}\ ]$

(i) Volume $=6a^2-24b^2$ cubic units

$6(a-2b)(a+2b)$

Answer$[\ 6,\;(a-2b),\;(a+2b)\ ]$

(ii) Volume $=3ps^2-15ps+12p$ cubic units

$3p(s^2-5s+4)=3p(s-1)(s-4)$

Answer$[\ 3p,\;(s-4),\;(s-1)\ ]$

Outer square side $=40+2s$
Area of path $=(40+2s)^2-40^2=(80+2s)(2s)=4s(40+s)$ sq. m
Expansion: $4s^2+160s$ sq. m

Answer$[\ 4s(s+40)\text{ sq. m}\ ]$

$x+\dfrac{1}{x}=\dfrac{10}{3}$ → $3x^2-10x+3=0$
Factorise: $(3x-1)(x-3)=0$
$x=3$ or $x=\dfrac{1}{3}$

Answer$\left[\ x=3\;\text{ or }\;x=\dfrac{1}{3}\ \right]$

$2x^2+7x+3=(2x+1)(x+3)$
Length $=\dfrac{(2x+1)(x+3)}{2x+1}=x+3$

Answer$[\ (x+3)\text{ hastas}\ ]$

Substitute $x=2$: $\;4p+10+r=0\quad\cdots(1)$
Substitute $x=\tfrac{1}{2}$, multiply by 4: $\;p+10+4r=0\quad\cdots(2)$
Subtract (2) from (1): $3p-3r=0 \;\Rightarrow\; p=r$ $\quad\blacksquare$

Answer$[\ p=r\;\text{(Proved)}\ ]$

$(a+b+c)^2=25\Rightarrow a^2+b^2+c^2+20=25\Rightarrow a^2+b^2+c^2=5$

$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
$=(5)(5-10)=(5)(-5)=-25\quad\blacksquare$

Answer$[\ -25\;\text{(Proved)}\ ]$

$n^3-n=n(n^2-1)=(n-1)\cdot n\cdot(n+1)$

This is a product of 3 consecutive integers.
∴ Divisible by 2 (one even factor) and by 3 (one multiple of 3).
Since gcd(2, 3) = 1, the product is divisible by 6. $\quad\blacksquare$

Answer$[\ (n-1)n(n+1)\text{ always divisible by }6\ ]$

(i) $x^3+y^3-12xy+64$, when $x+y=-4$

$=x^3+y^3+4^3-3(x)(y)(4)$; identity with $a=x,b=y,c=4$
$a+b+c=x+y+4=-4+4=0$
∴ Expression $=0$

Answer$[\ 0\ ]$

(ii) $x^3-8y^3-36xy-216$, when $x=2y+6$

$=x^3+(-2y)^3+(-6)^3-3(x)(-2y)(-6)$; identity with $a=x,b=-2y,c=-6$
$a+b+c=x-2y-6=0$ (given)
∴ Expression $=0$