What is factor 16d^8-8d^4+1 ?

The solution to factor 16d^8-8d^4+1 is (2d^2+1)^2(sqrt(2)d+1)^2(sqrt(2)d-1)^2

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factor 16d^8-8d^4+1

factor

16 d^{8} - 8 d^{4} + 1

(2 d^{2} + 1)^{2} (2 d + 1)^{2} (2 d - 1)^{2}

Solution steps

16 d^{8} - 8 d^{4} + 1

Let

u = d^{2}

= 16 u^{4} - 8 u^{2} + 1

Factor

16 u^{4} - 8 u^{2} + 1 : (2 u + 1) (2 u - 1) (2 u + 1) (2 u - 1)

= (2 u + 1) (2 u - 1) (2 u + 1) (2 u - 1)

Substitute back

u = d^{2}

= (2 d^{2} + 1) (2 d^{2} - 1) (2 d^{2} + 1) (2 d^{2} - 1)

Refine

= (2 d^{2} + 1)^{2} (2 d^{2} - 1)^{2}

Apply exponent rule:

a^{m} b^{m} = (ab)^{m},

assuming

a \geq 0, b \geq 0

(2 d^{2} + 1)^{2} (2 d^{2} - 1)^{2} = ((2 d^{2} + 1) (2 d^{2} - 1))^{2}

= ((2 d^{2} + 1) (2 d^{2} - 1))^{2}

Factor

(2 d^{2} + 1) (2 d^{2} - 1) : (2 d^{2} + 1) (2 d + 1) (2 d - 1)

= ((2 d^{2} + 1) (2 d + 1) (2 d - 1))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= (2 d^{2} + 1)^{2} (2 d + 1)^{2} (2 d - 1)^{2}

24 x + 18 y + 39

z + 4 (2 z + 3) = 15

\frac{9}{2} (2 + x) \leq \frac{1}{3} (5 x + 14) \leq 11 - \frac{3 x}{2}

- 3 r^{2} - 2 r - 5 = 0

6 x y^{0}

What is factor 16d^8-8d^4+1 ?
The solution to factor 16d^8-8d^4+1 is (2d^2+1)^2(sqrt(2)d+1)^2(sqrt(2)d-1)^2