How to factor 16x^916x9 + 54y^354y3 using difference of cubes?

1 Answer
Shwetank Mauria
Oct 10, 2017

16x^9+54y^3=2(2x^3+3y)(4x^4-6x^3y+9y^2)16x9+54y3=2(2x3+3y)(4x4−6x3y+9y2)

Explanation:

Here we have sum of cubes and not difference of cubes. Hence, we have to use a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2). As such

16x^9+54y^316x9+54y3

= 2(8x^9+27y^3)2(8x9+27y3)

= 2((2x^3)^3+(3y)^3)2((2x3)3+(3y)3)

= 2((2x^3)+(3y))((2x^3)^2-(2x^3)(3y)+(3y)^2)2((2x3)+(3y))((2x3)2−(2x3)(3y)+(3y)2)

= 2(2x^3+3y)(4x^4-6x^3y+9y^2)2(2x3+3y)(4x4−6x3y+9y2)

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