ossidianaZ

2021-09-16

The solution set of the equation ${\left(2y-3\right)}^{\frac{1}{3}}-{\left(4y+5\right)}^{\frac{1}{3}}=0$

Sally Cresswell

Step 1
Consider the provided equation,
${\left(2y-3\right)}^{\frac{1}{3}}-{\left(4y+5\right)}^{\frac{1}{3}}=0$
Now, add ${\left(4{y}_{5}\right)}^{\frac{1}{3}}$ to the both sides,
${\left(2y-3\right)}^{\frac{1}{3}}-{\left(4y+5\right)}^{\frac{1}{3}}=0$
${\left(2t-3\right)}^{\frac{1}{3}}-{\left(4y+5\right)}^{\frac{1}{3}}+{\left(4y+5\right)}^{\frac{1}{3}}={\left(4y+5\right)}^{\frac{1}{3}}$
${\left(2y-3\right)}^{\frac{1}{3}}={\left(4y+5\right)}^{\frac{1}{3}}$
Take cube on both sides of the above equation,
${\left(2y-3\right)}^{\frac{1}{3}}={\left(4y+5\right)}^{\frac{1}{3}}$
${\left({\left(2y-3\right)}^{\frac{1}{3}}\right)}^{3}={\left({\left(4y+5\right)}^{\frac{1}{3}}\right)}^{3}$
$2y-3=4y+5$
$-3-5=4y-2y$
Simplify the above equation,
$-3-5=4y-2y$
$-8=2y$
$y=-4$
Check:
Now, substitute the value of $y=-4$ in the provided equation $\left(2y-3{\right)}^{\frac{1}{3}}-\left(4y+5{\right)}^{\frac{1}{3}}=0$
$\left[2\left(-4\right)-3{\right]}^{\frac{1}{3}}-\left[4\left(-4\right)+5{\right]}^{\frac{1}{3}}\begin{array}{c}?\\ =\end{array}0$
$\left(-8-3{\right)}^{\frac{1}{3}}-\left(-16+5{\right)}^{\frac{1}{3}}\begin{array}{c}?\\ =\end{array}0$
$\left(-11{\right)}^{\frac{1}{3}}-\left(-11{\right)}^{\frac{1}{3}}\begin{array}{c}?\\ =\end{array}0$
$0\begin{array}{c}?\\ =\end{array}0$
Which is true.
Hence, the solution set of the equation ${\left(2y-3\right)}^{\frac{1}{3}}-{\left(4y+5\right)}^{\frac{1}{3}}=0$ is $\left\{-4\right\}$

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