Factor 8+27y^6

Math
8+27y6
Rewrite 8 as 23.
23+27y6
Rewrite 27y6 as (3y2)3.
23+(3y2)3
Since both terms are perfect cubes, factor using the sum of cubes formula, a3+b3=(a+b)(a2-ab+b2) where a=2 and b=3y2.
(2+3y2)(22-2(3y2)+(3y2)2)
Simplify.
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Raise 2 to the power of 2.
(2+3y2)(4-2(3y2)+(3y2)2)
Multiply 3 by -2.
(2+3y2)(4-6y2+(3y2)2)
Apply the product rule to 3y2.
(2+3y2)(4-6y2+32(y2)2)
Raise 3 to the power of 2.
(2+3y2)(4-6y2+9(y2)2)
Multiply the exponents in (y2)2.
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Apply the power rule and multiply exponents, (am)n=amn.
(2+3y2)(4-6y2+9y2⋅2)
Multiply 2 by 2.
(2+3y2)(4-6y2+9y4)
(2+3y2)(4-6y2+9y4)
(2+3y2)(4-6y2+9y4)
Factor 8+27y^6

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