Question
Verify : $x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
✓
Answer
We know that
$(x-y)^3=x^3-y^3-3 x y(x-y)\left\{\text { Using Identity }(a-b)^3=a^3-b^3-3 a b(a-b)\right\}$
$\Rightarrow x^3-y^3=(x-y)^3+3 x y(x-y)$
$\Rightarrow x^3-y^3=(x-y)\left\{(x-y)^2+3 x y\right\}$
$\Rightarrow x^3-y^3=(x-y)\left(x^2-2 x y+y^2+3 x y\right)$
$\Rightarrow x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
$(x-y)^3=x^3-y^3-3 x y(x-y)\left\{\text { Using Identity }(a-b)^3=a^3-b^3-3 a b(a-b)\right\}$
$\Rightarrow x^3-y^3=(x-y)^3+3 x y(x-y)$
$\Rightarrow x^3-y^3=(x-y)\left\{(x-y)^2+3 x y\right\}$
$\Rightarrow x^3-y^3=(x-y)\left(x^2-2 x y+y^2+3 x y\right)$
$\Rightarrow x^3-y^3=(x-y)\left(x^2+x y+y^2\right)$
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$8 x^3+y^3+12 x^2 y+6 x y^2$
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