What is factor b^9-1 ?

The solution to factor b^9-1 is (b-1)(b^2+b+1)(b^6+b^3+1)

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factor b^9-1

factor

b^{9} - 1

(b - 1) (b^{2} + b + 1) (b^{6} + b^{3} + 1)

Solution steps

b^{9} - 1

Apply exponent rules

= (b^{3})^{3} - 1^{3}

Apply Difference of Cubes Formula:

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

(b^{3})^{3} - 1^{3} = (b^{3} - 1) (b^{6} + b^{3} + 1)

= (b^{3} - 1) (b^{6} + b^{3} + 1)

Factor

b^{3} - 1 : (b - 1) (b^{2} + b + 1)

= (b - 1) (b^{2} + b + 1) (b^{6} + b^{3} + 1)

5 (g - 3) > 7 (2 g + 5) - (- 4 g)

3 x^{2} + 18 x + 30

5 + (- 3) - 6 - (- 1)

- 32 m^{4} n^{2} + 14 m n^{3}

What is factor b^9-1 ?
The solution to factor b^9-1 is (b-1)(b^2+b+1)(b^6+b^3+1)