Question

Solve the following mathematical equation: 

\( (3x +1)^{\frac{3}{2}} - 24 = 60 \)

Answer 1

To find the value of \( x \), we will assume:

\( 3x + 1 = u \)

Now, substituting \( 3x + 1 = u \) into the given equation, we have:

\( u^{\frac{3}{2}} - 24 = 40 \)

\( \Rightarrow u^{\frac{3}{2}} = 40 + 24 \)

\( \Rightarrow u^{\frac{3}{2}} = 64 \)

Squaring both sides, we have:

\( \Rightarrow \left( u^{\frac{3}{2}} \right)^2 = (64)^2 \)

\( \Rightarrow u^3 = (64)^2 \)

\( \Rightarrow u = \left( 64 \right)^{\frac{2}{3}} \, \, \, \, \left[ \because a^{n} = b \,\, \therefore a = b^{\frac{1}{n}} \right] \)

\( \Rightarrow u = (4^3)^{\frac{2}{3}} \)

\( \Rightarrow u = 4^2 \)

\( \Rightarrow u = 16 \)

Now, undo substitution \( u = 16 \) into \( 3x + 1 = u \) , we have:

\( 3x + 1 = 16 \)

\( \Rightarrow 3x = 16 - 1 \)

\( \Rightarrow x = \dfrac{15}{3} \)

\( \Rightarrow x = 5 \)

 

Hence, the solution of the given equation is \( x = \color{blue}{\boxed{ 5 }} \)