What is expand 7/((x+3)(2x-1)(3x+2)) ?

The solution to expand 7/((x+3)(2x-1)(3x+2)) is 7/(6x^3+19x^2+x-6)

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expand 7/((x+3)(2x-1)(3x+2))

expand

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

\frac{7}{6 x ^{3} + 19 x ^{2} + x - 6}

Solution steps

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

Expand

(x + 3) (2 x - 1) (3 x + 2) : 6 x^{3} + 19 x^{2} + x - 6

= \frac{7}{6 x ^{3} + 19 x ^{2} + x - 6}

π r (2 h + r)

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

\frac{1}{x ^{2} ( x - 1 )}

\frac{1}{2} (5 ln (5) - 4)

(x + \frac{2}{x ^{2}})^{15}

What is expand 7/((x+3)(2x-1)(3x+2)) ?
The solution to expand 7/((x+3)(2x-1)(3x+2)) is 7/(6x^3+19x^2+x-6)