What is factor 64m^{12}+27n^6 ?

The solution to factor 64m^{12}+27n^6 is (4m^4+3n^2)(4m^4+3n^2+6m^2n)(4m^4+3n^2-6m^2n)

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factor 64m^{12}+27n^6

factor

64 m^{12} + 27 n^{6}

(4 m^{4} + 3 n^{2}) (4 m^{4} + 3 n^{2} + 6 m^{2} n) (4 m^{4} + 3 n^{2} - 6 m^{2} n)

Solution steps

64 m^{12} + 27 n^{6}

Apply exponent rules

= (4 m^{4})^{3} + (3 n^{2})^{3}

Apply Sum of Cubes Formula:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(4 m^{4})^{3} + (3 n^{2})^{3} = (4 m^{4} + 3 n^{2}) ((4 m^{4})^{2} - 4 m^{4} \cdot 3 n^{2} + (3 n^{2})^{2})

= (4 m^{4} + 3 n^{2}) ((4 m^{4})^{2} - 4 m^{4} \cdot 3 n^{2} + (3 n^{2})^{2})

Simplify

(4 m^{4})^{2} - 4 m^{4} \cdot 3 n^{2} + (3 n^{2})^{2} : 16 m^{8} - 12 m^{4} n^{2} + 9 n^{4}

= (4 m^{4} + 3 n^{2}) (16 m^{8} - 12 m^{4} n^{2} + 9 n^{4})

Factor

16 m^{8} - 12 m^{4} n^{2} + 9 n^{4} : (4 m^{4} + 3 n^{2} + 6 m^{2} n) (4 m^{4} + 3 n^{2} - 6 m^{2} n)

= (4 m^{4} + 3 n^{2}) (4 m^{4} + 3 n^{2} + 6 m^{2} n) (4 m^{4} + 3 n^{2} - 6 m^{2} n)

(\frac{2}{3 x})^{4}

\frac{tan ^{2} ( x )}{sec ( x ) - 1}

20 f - 4 (5 f + 4) + 16

\frac{( x + 1 - 1 )}{x}

\frac{1 - x}{1 - x}

What is factor 64m^{12}+27n^6 ?
The solution to factor 64m^{12}+27n^6 is (4m^4+3n^2)(4m^4+3n^2+6m^2n)(4m^4+3n^2-6m^2n)