Question

solve it fast plz..​

Answer 1

Step-by-step explanation:

$Given:(\frac{3}{2} x+ 1)^3$

$=>(\frac{3x}{2} + 1)^3$

∴ (a + b)³ = a³ + b³ + 3a²b + 3ab²

$=>(\frac{3x}{2})^3 + (1)^3 + 3(\frac{3x}{2})^2(1) + 3(\frac{3x}{2})(1)^2$

$=>\boxed{ \frac{27x^3}{8}+1+\frac{27x^2}{4}+\frac{9x}{2}}$

Hope it helps!

Answer 2

$\huge\bf\pink{\mid{\overline{\underline{Your\: Answer}}}\mid}$

27${x}^{3}$/8 + 1 + 27${x}^{2}$/4 + 9x/2

step-by-step explanation:

A.T.Q.,

We have to find the value of,

${(3x/2+1)}^{3}$

Now,

we know that,

${(a+b)}^{3}$= ${a}^{3}$+${b}^{3}$+3${a}^{2}$b+3a${b}^{2}$

Observing this,

a = 3x/2

b = 1

putting the value of a and b in the formula,

we get,

${(3x/2+1)}^{3}$

= ${(3x/2)}^{3}$+${1}^{3}$+3${(3x/2)}^{2}$×1 + 3(3x/2)${1}^{2}$

= 27${x}^{3}$/8 + 1 + 3×(9${x}^{2}$/4) + 3(3x/2)

= 27${x}^{3}$/8 + 1 + 27${x}^{2}$/4 + 9x/2