What is factor 64j^6-k^9 ?

The solution to factor 64j^6-k^9 is (4j^2-k^3)(16j^4+4j^2k^3+k^6)

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factor 64j^6-k^9

factor

64 j^{6} - k^{9}

(4 j^{2} - k^{3}) (16 j^{4} + 4 j^{2} k^{3} + k^{6})

Solution steps

64 j^{6} - k^{9}

Apply exponent rules

= (4 j^{2})^{3} - (k^{3})^{3}

Apply Difference of Cubes Formula:

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

(4 j^{2})^{3} - (k^{3})^{3} = (4 j^{2} - k^{3}) ((4 j^{2})^{2} + 4 j^{2} k^{3} + (k^{3})^{2})

= (4 j^{2} - k^{3}) ((4 j^{2})^{2} + 4 j^{2} k^{3} + (k^{3})^{2})

Simplify

(4 j^{2})^{2} + 4 j^{2} k^{3} + (k^{3})^{2} : 16 j^{4} + 4 j^{2} k^{3} + k^{6}

= (4 j^{2} - k^{3}) (16 j^{4} + 4 j^{2} k^{3} + k^{6})

20 \div \frac{1}{2}

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

24 x^{3} y

\frac{4 x ^{3} + x + 1}{2 x ^{2} - x + 1}

49 x^{4} - 4 y^{2}

What is factor 64j^6-k^9 ?
The solution to factor 64j^6-k^9 is (4j^2-k^3)(16j^4+4j^2k^3+k^6)