Question

Without actual calculating, find:

$(x-2y{)}^3+(2y-32{)}^3+(32-x{)}^3-3((x-2y)(2y-32)(32-x))$.

Answer 1

We know the identity of trinomial: $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

So, comparing the given equation $(x-2y{)}^3+(2y-32{)}^3+(32-x{)}^3-3((x-2y)(2y-32)(32-x))$ with the identity $a^3+b^3+c^3-3abc$, we have,

$a=x-2y,b=2y-32,c=32-x$.

So, substitute the values of $a,b$ and $c$ in $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ as follows:

$(x-2y{)}^3+(2y-32{)}^3+(32-x{)}^3-3((x-2y)(2y-32)(32-x))$

$=(x-2y+2y-32+32-x)\left(\right(x-2y{)}^2+(2y-32{)}^2+(32-x{)}^2-(x-2y)(2y-32)$

$-(2y-32)(32-x)-(32-x)(x-2y))=0\times ((x-2y{)}^2+(2y-32{)}^2+(32-x{)}^2-(x-2y\left)\right(2y-32)-(2y-32\left)\right(32-x)-(32-x\left)\right(x-2y\left)\right)=0.$

Hence, $(x-2y{)}^3+(2y-32{)}^3+(32-x{)}^3-3((x-2y)(2y-32)(32-x))=0$.