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

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

= \frac{8 a ^{6} - 8 a ^{4} + 4 a ^{2} - 1}{8}

Apply the fraction rule:

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

\frac{8 a ^{6} - 8 a ^{4} + 4 a ^{2} - 1}{8} = \frac{8 a ^{6}}{8} - \frac{8 a ^{4}}{8} + \frac{4 a ^{2}}{8} - \frac{1}{8}

= \frac{8 a ^{6}}{8} - \frac{8 a ^{4}}{8} + \frac{4 a ^{2}}{8} - \frac{1}{8}

Divide the numbers:

\frac{8}{8} = 1

= a^{6} - a^{4} + \frac{4 a ^{2}}{8} - \frac{1}{8}

Cancel

\frac{4 a ^{2}}{8} : \frac{a ^{2}}{2}

= a^{6} - a^{4} + \frac{a ^{2}}{2} - \frac{1}{8}

- \frac{- l ( - 2 l ( 3 a - l ^{2} ) - a ) + 3 a}{2}

(y^{2} e^{x y^{2}} + 4 x^{3}) d x + (2 x y e^{x y^{2}} - 3 y^{2}) d y

\frac{x ^{4} + 4 x ^{2} - 9}{4 x ( x ^{2} + 6 x - 9 )}

\frac{( 4 x ^{2} + 5 )}{( x ^{2} + 1 ) ( x ^{2} + 9 )}

\frac{3}{x ^{2} - 2 x + 1} + \frac{2}{x ^{2} - 1}

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