What is factor b^7+128 ?

The solution to factor b^7+128 is (b+2)(b^6-2b^5+4b^4-8b^3+16b^2-32b+64)

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factor b^7+128

factor

b^{7} + 128

(b + 2) (b^{6} - 2 b^{5} + 4 b^{4} - 8 b^{3} + 16 b^{2} - 32 b + 64)

Solution steps

b^{7} + 128

Rewrite

128

2^{7}

= b^{7} + 2^{7}

Apply factoring rule:

x^{n} + y^{n} = (x + y) (x^{n - 1} - x^{n - 2} y + \dots - x y^{n - 2} + y^{n - 1}), n i s o dd

b^{7} + 2^{7} = (b + 2) (b^{6} - 2 b^{5} + 2^{2} b^{4} - 2^{3} b^{3} + 2^{4} b^{2} - 2^{5} b + 2^{6})

= (b + 2) (b^{6} - 2 b^{5} + 2^{2} b^{4} - 2^{3} b^{3} + 2^{4} b^{2} - 2^{5} b + 2^{6})

Simplify

(b + 2) (b^{6} - 2 b^{5} + 2^{2} b^{4} - 2^{3} b^{3} + 2^{4} b^{2} - 2^{5} b + 2^{6}) : (b + 2) (b^{6} - 2 b^{5} + 4 b^{4} - 8 b^{3} + 16 b^{2} - 32 b + 64)

= (b + 2) (b^{6} - 2 b^{5} + 4 b^{4} - 8 b^{3} + 16 b^{2} - 32 b + 64)

\frac{1 - 3}{2 2}

(2 p - 1) (2 p + 1)

x^{4} - 8 x^{3} + 24 x^{2} + 8 x - 25

1.027 \cdot 50 \cdot 0.001

1 \frac{3}{4} \times \frac{1}{2}

What is factor b^7+128 ?
The solution to factor b^7+128 is (b+2)(b^6-2b^5+4b^4-8b^3+16b^2-32b+64)