Question

What is 125x9 + 64y12 written as a sum of cubes?

Answer 1

`\bf \textit{difference and sum of cubes} \\\\ a^3+b^3 = (a+b)(a^2-ab+b^2) \\\\ a^3-b^3 = (a-b)(a^2+ab+b^2)\\\\ -------------------------------`


`\bf 125x^9+64y^(12)\qquad \begin{cases} 125=5\cdot 5\cdot 5\\ \qquad 5^3\\ x^9=x^(3\cdot 3)\\ \qquad (x^3)^3\\ 64=4\cdot 4\cdot 4\\ \qquad 4^3\\ y^(12)=y^(4\cdot 3)\\ \qquad (y^4)^3 \end{cases}\implies 5^3(x^3)^3+4^3(y^4)^3 \\\\\\ (5x^3)^3+(4y^4)^3\implies [(5x^3)^2-(5x^3)(4y^4)+(4y^4)^2] \\\\\\ 25x^6-20x^3y^4+16y^8`

Answer 2

`125x^(9) +64y^(12)= (5x^(3) +4y^(4) )(25x^(6) -20x^(3) y^(4) +16y^(8))`

Explanation:

The sum of cubes can be factored with the following notable product

`(x^(3) +y^(3) )=(x+y)(x^(2) -xy+y^(2) )`

So, in this case, we have:

`x=5x^(3)`

`y=4y^(4)`

replacing

`125x^(9)+64y^(12) =(5x^(3)+4y^(4)  )((5x^(3))^(2)-(5x^(3) )(4y^(4))+(4y^(4))^(2))`

performing operations

`125x^(9) +64y^(12) = (5x^(3) +4y^(4) )(25x^(6) -20x^(3) y^(4) +16y^(8))`