What is factor 27a^{3m}+64b^{6n} ?

The solution to factor 27a^{3m}+64b^{6n} is (3a^m+4b^{2n})(9a^{2m}-12a^mb^{2n}+16b^{4n})

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factor 27a^{3m}+64b^{6n}

factor

27 a^{3 m} + 64 b^{6 n}

(3 a^{m} + 4 b^{2 n}) (9 a^{2 m} - 12 a^{m} b^{2 n} + 16 b^{4 n})

Solution steps

27 a^{3 m} + 64 b^{6 n}

Rewrite

27 a^{3 m}

(3 a^{1 m})^{3}

Rewrite

64 b^{6 n}

(2 b^{1 n})^{6}

= (3 a^{1 \cdot m})^{3} + (2 b^{1 \cdot n})^{6}

Apply exponent rule:

a^{b c} = (a^{b})^{c}

(2 b^{1 n})^{6} = ((2 b^{1 n})^{2})^{3}

= (3 a^{1 m})^{3} + ((2 b^{1 n})^{2})^{3}

Apply Sum of Cubes Formula:

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

(3 a^{1 m})^{3} + ((2 b^{1 n})^{2})^{3} = (3 a^{1 m} + (2 b^{1 n})^{2}) ((3 a^{1 m})^{2} - 3 a^{1 m} (2 b^{1 n})^{2} + (2 b^{1 n})^{4})

= (3 a^{1 m} + (2 b^{1 n})^{2}) ((3 a^{1 m})^{2} - 3 a^{1 m} (2 b^{1 n})^{2} + (2 b^{1 n})^{4})

Simplify

(3 a^{1 \cdot m} + (2 b^{1 \cdot n})^{2}) ((3 a^{1 \cdot m})^{2} - 3 a^{1 \cdot m} (2 b^{1 \cdot n})^{2} + (2 b^{1 \cdot n})^{4}) : (3 a^{m} + 4 b^{2 n}) (9 a^{2 m} - 12 a^{m} b^{2 n} + 16 b^{4 n})

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

\frac{d ^{2}}{d x ^{2}} ((x) (2 C x))

144 x^{(2)} - y^{(2)} - 160 x - 5 y

A (- 7.3)

\frac{( 3 x ^{2} - 16 )}{8 - x}

\frac{9. 1 ^{2} - 3}{9.1 - 9}

What is factor 27a^{3m}+64b^{6n} ?
The solution to factor 27a^{3m}+64b^{6n} is (3a^m+4b^{2n})(9a^{2m}-12a^mb^{2n}+16b^{4n})