Question

Shew that $(x^2+6x^{2}y+3x{y}^2-y^3{)}^3+(y^2+6x{y}^2+3x^{2}y-x^3{)}^3=27xy(x+y)(x^2+xy+y^2{)}^3$.

Answer 1

We know $A^3+B^3=(A+B)(A+wB)(A+w^{2}B)$

On putting, $A=x^3+6x^{2}y+3x{y}^2-y^3$

$B=y^3+6x{y}^2+3x^{2}y-x^3$

We obtain,

$A+B=9x^{2}y+9x{y}^2=9xy(x+y)$

Also,

$A+wB=x^3(1-w)+3w^{2}y(2+w)+3x{y}^2(1+2w)-y^3(1-w)$

$=(1-w)(x^3-3w^2{x}^{2}y+3w^{4}x{y}^2-w^6{y}^3)$

For,

$2+w=1+w+w^3=-w^2+w^3=-w^2(1-w)$

and $(1+2w)=-(w+w^2)+2w=w(1-w)$

$=w^4(1-w)$

Thus, $A+wB=(1-w){(x-w^{2}y)}^3$

By writing $w^2$ for 'w' and 'w' for $w^2$, we have

$A+w^{2}B=(1-w^2){(x-wy)}^3$

$\therefore A^3+B^3=9xy(x+y)(1-w)(1-w^2){(x-wy)}^3{(x-w^{2}y)}^3$

$=27xy(x+y){(x^2+xy+y^2)}^3$

$\because w^3=1$ and $w+w^2=-1$