What is expand-(sqrt(8-2x^2)(4-x^2))/3 ?

The solution to expand-(sqrt(8-2x^2)(4-x^2))/3 is -(4sqrt(8-2x^2))/3+(x^2sqrt(8-2x^2))/3

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expand-(sqrt(8-2x^2)(4-x^2))/3

expand

- \frac{8 - 2 x ^{2} ( 4 - x ^{2} )}{3}

- \frac{4 8 - 2 x ^{2}}{3} + \frac{x ^{2} 8 - 2 x ^{2}}{3}

Solution steps

- \frac{8 - 2 x ^{2} ( 4 - x ^{2} )}{3}

Expand

8 - 2 x^{2} (4 - x^{2}) : 4 8 - 2 x^{2} - x^{2} 8 - 2 x^{2}

= - \frac{- x ^{2} - 2 x ^{2} + 8 + 4 - 2 x ^{2} + 8}{3}

Apply the fraction rule:

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

\frac{4 8 - 2 x ^{2} - x ^{2} 8 - 2 x ^{2}}{3} = - (\frac{4 8 - 2 x ^{2}}{3}) - (- \frac{x ^{2} 8 - 2 x ^{2}}{3})

= - (\frac{4 - 2 x ^{2} + 8}{3}) - (- \frac{x ^{2} - 2 x ^{2} + 8}{3})

Remove parentheses:

(a) = a, - (- a) = a

= - \frac{4 8 - 2 x ^{2}}{3} + \frac{x ^{2} 8 - 2 x ^{2}}{3}

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

(\frac{2}{3} x^{6} + \frac{1}{4} y) (\frac{2}{3} x^{6} - \frac{1}{4} y)

3 l e f t (c - 6 b + 6 z)

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

(\frac{1}{2} a + \frac{2}{3} a^{\frac{4}{5}})^{12}

What is expand-(sqrt(8-2x^2)(4-x^2))/3 ?
The solution to expand-(sqrt(8-2x^2)(4-x^2))/3 is -(4sqrt(8-2x^2))/3+(x^2sqrt(8-2x^2))/3