What is expand ((-2a^2-1)(4a^4-2a^2+1))/8 ?

The solution to expand ((-2a^2-1)(4a^4-2a^2+1))/8 is -a^6-1/8

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expand ((-2a^2-1)(4a^4-2a^2+1))/8

expand

\frac{( - 2 a ^{2} - 1 ) ( 4 a ^{4} - 2 a ^{2} + 1 )}{8}

- a^{6} - \frac{1}{8}

Solution steps

\frac{( - 2 a ^{2} - 1 ) ( 4 a ^{4} - 2 a ^{2} + 1 )}{8}

Expand

(- 2 a^{2} - 1) (4 a^{4} - 2 a^{2} + 1) : - 8 a^{6} - 1

= \frac{- 8 a ^{6} - 1}{8}

Apply the fraction rule:

\frac{a \pm b}{c} = \frac{a}{c} \pm \frac{b}{c}

\frac{- 8 a ^{6} - 1}{8} = - \frac{8 a ^{6}}{8} - \frac{1}{8}

= - \frac{8 a ^{6}}{8} - \frac{1}{8}

Divide the numbers:

\frac{8}{8} = 1

= - a^{6} - \frac{1}{8}

\frac{i}{( a + bi ) ^{2}}

\frac{2 x ^{3} + 5 x - 1}{( 4 x + 7 ) ^{2} ( x ^{2} + 1 ) ^{3}}

(x^{2} + e^{4 x} - 7)^{2}

ro t (x^{2} y, 1 + y, x z^{2})

\frac{2 ( 1 + x ) ^{3}}{3}

What is expand ((-2a^2-1)(4a^4-2a^2+1))/8 ?
The solution to expand ((-2a^2-1)(4a^4-2a^2+1))/8 is -a^6-1/8