Write the following cube in expanded form, 32x+13 .

# Write the following cube in expanded form, ${\left(\frac{3}{2}x+1\right)}^{3}$ .

1. A
$\frac{27}{8}{x}^{3}+\frac{27}{4}{x}^{2}-\frac{9}{2}x+1$
2. B
$\frac{27}{8}{x}^{3}+\frac{27}{4}{x}^{2}+\frac{9}{2}x+1$
3. C
$\frac{27}{8}{x}^{3}+\frac{27}{4}{x}^{2}+\frac{9}{2}x-1$
4. D
$\frac{27}{8}{x}^{3}+\frac{27}{4}{x}^{2}-\frac{9}{2}x-1$

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### Solution:

It is given ${\left(\frac{3}{2}x+1\right)}^{3}.$
We know the algebraic identity,  ${\left(a+b\right)}^{3}={a}^{3}+{b}^{3}+3\mathit{ab}\left(a+b\right)$.
Applying identity ${\left(a+b\right)}^{3}={a}^{3}+{b}^{3}+3\mathit{ab}\left(a+b\right)$ in ${\left(\frac{3}{2}x+1\right)}^{3}$, where, $a=\frac{3}{2}$x, and b = 1, we get,  ${\left(\frac{3}{2}x+1\right)}^{3}={\left(\frac{3}{2}x\right)}^{3}+{1}^{3}+3\left(\frac{3}{2}x\right)\left(1\right)\left(\frac{3}{2}x+1\right)$
${\left(\frac{3}{2}x+1\right)}^{3}=\frac{27}{8}{x}^{3}+1+\frac{9}{2}x\left(\frac{3}{2}x+1\right)$
${\left(\frac{3}{2}x+1\right)}^{3}=\frac{27}{8}{x}^{3}+1+\left(\frac{9}{2}x\right)×\left(\frac{3}{2}x\right)+\left(\frac{9}{2}x\right)$
${\left(\frac{3}{2}x+1\right)}^{3}=\frac{27}{8}{x}^{3}+1+\frac{27}{4}{x}^{2}+\frac{9}{2}x$
${\left(\frac{3}{2}x+1\right)}^{3}=\frac{27}{8}{x}^{3}+\frac{27}{4}{x}^{2}+\frac{9}{2}x+1$
Therefore, option 2 is  correct.

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