Question

Without actually calculating the cubes, find the values of ${(-10)}^3+3^3+7^3$

Answer 1

We know the corollary: if $a+b+c=0$ then $a^3+b^3+c^3=3abc$

Using the above corollary taking $a=-10$, $b=3$ and $c=7$, we have $a+b+c=-10+3+7=0$ then the value of $(-10{)}^3+(3{)}^3+(7{)}^3$ is:

$(-10{)}^3+(3{)}^3+(7{)}^3=3(-10\times 3\times 7)=-630$

Hence, $(-10{)}^3+(3{)}^3+(7{)}^3=-630$