Factor 27v^6+64g^3

Math
27v6+64g3
Rewrite 27v6 as (3v2)3.
(3v2)3+64g3
Rewrite 64g3 as (4g)3.
(3v2)3+(4g)3
Since both terms are perfect cubes, factor using the sum of cubes formula, a3+b3=(a+b)(a2-ab+b2) where a=3v2 and b=4g.
(3v2+4g)((3v2)2-(3v2)(4g)+(4g)2)
Simplify.
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Apply the product rule to 3v2.
(3v2+4g)(32(v2)2-(3v2)(4g)+(4g)2)
Raise 3 to the power of 2.
(3v2+4g)(9(v2)2-(3v2)(4g)+(4g)2)
Multiply the exponents in (v2)2.
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Apply the power rule and multiply exponents, (am)n=amn.
(3v2+4g)(9v2⋅2-(3v2)(4g)+(4g)2)
Multiply 2 by 2.
(3v2+4g)(9v4-(3v2)(4g)+(4g)2)
(3v2+4g)(9v4-(3v2)(4g)+(4g)2)
Multiply 3 by -1.
(3v2+4g)(9v4-3v2(4g)+(4g)2)
Rewrite using the commutative property of multiplication.
(3v2+4g)(9v4-3⋅4v2g+(4g)2)
Multiply -3 by 4.
(3v2+4g)(9v4-12v2g+(4g)2)
Apply the product rule to 4g.
(3v2+4g)(9v4-12v2g+42g2)
Raise 4 to the power of 2.
(3v2+4g)(9v4-12v2g+16g2)
(3v2+4g)(9v4-12v2g+16g2)
Factor 27v^6+64g^3

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