What is expand (ir(2+x^2)^3)/8 ?

The solution to expand (ir(2+x^2)^3)/8 is i(8r+rx^6+12rx^2+6rx^4)/8

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expand (ir(2+x^2)^3)/8

expand

\frac{i r ( 2 + x ^{2} ) ^{3}}{8}

i \frac{8 r + r x ^{6} + 12 r x ^{2} + 6 r x ^{4}}{8}

Solution steps

\frac{i r ( 2 + x ^{2} ) ^{3}}{8}

(2 + x^{2})^{3} = 8 + 12 x^{2} + 6 x^{4} + x^{6}

= \frac{i r ( x ^{6} + 6 x ^{4} + 12 x ^{2} + 8 )}{8}

Expand

i r (8 + 12 x^{2} + 6 x^{4} + x^{6}) : 8 i r + 12 i r x^{2} + 6 i r x^{4} + i r x^{6}

= \frac{8 i r + 12 i r x ^{2} + 6 i r x ^{4} + i r x ^{6}}{8}

Apply the fraction rule:

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

\frac{8 i r + 12 i r x ^{2} + 6 i r x ^{4} + i r x ^{6}}{8} = \frac{8 i r}{8} + \frac{12 i r x ^{2}}{8} + \frac{6 i r x ^{4}}{8} + \frac{i r x ^{6}}{8}

= \frac{8 i r}{8} + \frac{12 i r x ^{2}}{8} + \frac{6 i r x ^{4}}{8} + \frac{i r x ^{6}}{8}

Divide the numbers:

\frac{8}{8} = 1

= i r + \frac{12 i r x ^{2}}{8} + \frac{6 i r x ^{4}}{8} + \frac{i r x ^{6}}{8}

Cancel

\frac{12 i r x ^{2}}{8} : \frac{3 i r x ^{2}}{2}

= i r + \frac{3 i r x ^{2}}{2} + \frac{6 i r x ^{4}}{8} + \frac{i r x ^{6}}{8}

Cancel

\frac{6 i r x ^{4}}{8} : \frac{3 i r x ^{4}}{4}

= i r + \frac{3 i r x ^{2}}{2} + \frac{3 i r x ^{4}}{4} + \frac{i r x ^{6}}{8}

Rewrite

i r + \frac{3 i r x ^{2}}{2} + \frac{3 i r x ^{4}}{4} + \frac{i r x ^{6}}{8}

in standard complex form:

\frac{8 r + 12 r x ^{2} + 6 r x ^{4} + r x ^{6}}{8} i

= \frac{8 r + 12 r x ^{2} + 6 r x ^{4} + r x ^{6}}{8} i

\frac{( 2 x - 3 ) ( x + 9 )}{4 x ( 2 x - 9 )}

\frac{- 2 s ^{2} - 1}{( s ^{2} + 1 ) ^{2}}

A (4, - 1) B (2, 3) C (2, 3)

\frac{6 s ^{2} + 50}{( s + 3 ) ( s + 4 )}

\frac{( ( s + 1 ) ( s + 2 ) )}{s ^{3} ( s + 3 ) ( s + 4 )}

What is expand (ir(2+x^2)^3)/8 ?
The solution to expand (ir(2+x^2)^3)/8 is i(8r+rx^6+12rx^2+6rx^4)/8