How do you factor the expression $81x^6+24x^3y^3$ ?

Question

2590 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-10T08:24:15

$81x^6 + 24x^3y^3 = 3x^3(3x+2y)(9x^2-6xy+4y^2)$

Given:

$81x^6 + 24x^3y^3$

Note that both factors are divisible by $3x^3$ . So we can separate out that factor first to find:

$81x^6 + 24x^3y^3 = 3x^3(27x^3+8y^3)$

Next note that both $27x^3 = (3x)^3$ and $8y^3 = (2y)^3$ are perfect cubes. So we can use the sum of cubes identity:

$A^3+B^3 = (A+B)(A^2-AB+B^2)$

with $A=3x$ and $B=2y$ to find:

$27x^3+8y^3 = (3x)^3+(2y)^3$

$color(white)(27x^3+8y^3) = (3x+2y)((3x)^2-(3x)(2y)+(2y)^2)$

$color(white)(27x^3+8y^3) = (3x+2y)(9x^2-6xy+4y^2)$

Putting it all together:

$81x^6 + 24x^3y^3 = 3x^3(3x+2y)(9x^2-6xy+4y^2)$

How do you factor the expression 81x^6+24x^3y^381x^6+24x^3y^3?