What is factor a^6-2a^3b^3+b^6 ?

The solution to factor a^6-2a^3b^3+b^6 is (a-b)^2(a^2+ab+b^2)^2

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factor a^6-2a^3b^3+b^6

factor

a^{6} - 2 a^{3} b^{3} + b^{6}

(a - b)^{2} (a^{2} + ab + b^{2})^{2}

Solution steps

a^{6} - 2 a^{3} b^{3} + b^{6}

Break the expression into groups

= (a^{6} - a^{3} b^{3}) + (- a^{3} b^{3} + b^{6})

Factor out

a^{3}

from

a^{6} - a^{3} b^{3} : a^{3} (a^{3} - b^{3})

Factor out

- b^{3}

from

- a^{3} b^{3} + b^{6} : - b^{3} (a^{3} - b^{3})

= a^{3} (a^{3} - b^{3}) - b^{3} (a^{3} - b^{3})

Factor out common term

a^{3} - b^{3}

= (a^{3} - b^{3}) (a^{3} - b^{3})

Factor

a^{3} - b^{3} : (a - b) (a^{2} + ab + b^{2})

= (a - b) (a^{2} + ab + b^{2}) (a^{3} - b^{3})

Factor

a^{3} - b^{3} : (a - b) (a^{2} + ab + b^{2})

= (a - b) (a^{2} + ab + b^{2}) (a - b) (a^{2} + ab + b^{2})

Simplify

= (a - b)^{2} (a^{2} + ab + b^{2})^{2}

\frac{1}{3} (4 \cdot 3) + 2^{3}

w^{2} - 64

8 x^{2} + 64 = 0

2 x^{2} + 2 x \geq x^{2} - 2 x + 5

24 x^{3} - 9 x^{2} y^{2} - 12

What is factor a^6-2a^3b^3+b^6 ?
The solution to factor a^6-2a^3b^3+b^6 is (a-b)^2(a^2+ab+b^2)^2