Question

If $\alpha =\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}$, then the value of ${{\alpha }^{3}}-21\alpha -90$?

Answer 1

Hint: We start solving the problem by assuming $\alpha ={{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}}$such that $x=45+29\sqrt{2}$, $\,y=45-29\sqrt{2}$. We then apply cubing on both sides of $\alpha ={{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}}$ using the fact that ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$ to find ${{\alpha }^{3}}$. We then make the necessary arrangements in ${{\alpha }^{3}}$ to proceed through the problem. We then find the value of ${{\left( xy \right)}^{\dfrac{1}{3}}}$ and substitute it in ${{\alpha }^{3}}$. We then also use $x=45+29\sqrt{2}$, $\,y=45-29\sqrt{2}$ and make necessary calculations to get the required answer.

Complete step-by-step answer:
According to the problem, we are given $\alpha =\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}$ and we need to find the value of ${{\alpha }^{3}}-21\alpha -90$.
Let us assume $\alpha ={{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}}$such that $x=45+29\sqrt{2}$, $\,y=45-29\sqrt{2}$.
Let us first find the value of ${{\alpha }^{3}}$.
So, we have $\alpha ={{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}}$.
Let us now apply cubing on both sides.
$\Rightarrow {{\alpha }^{3}}={{\left( {{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}} \right)}^{3}}$.
We know that ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$.
$\Rightarrow {{\alpha }^{3}}={{\left( {{x}^{\dfrac{1}{3}}} \right)}^{3}}+3{{\left( {{x}^{\dfrac{1}{3}}} \right)}^{2}}{{y}^{\dfrac{1}{3}}}+3{{x}^{\dfrac{1}{3}}}{{\left( {{y}^{\dfrac{1}{3}}} \right)}^{2}}+{{\left( {{y}^{\dfrac{1}{3}}} \right)}^{3}}$.
$\Rightarrow {{\alpha }^{3}}=x+3\left( {{x}^{\dfrac{1}{3}}}{{y}^{\dfrac{1}{3}}} \right){{x}^{\dfrac{1}{3}}}+3\left( {{x}^{\dfrac{1}{3}}}{{y}^{\dfrac{1}{3}}} \right){{y}^{\dfrac{1}{3}}}+y$.
On simplifying, we get
$\Rightarrow {{\alpha }^{3}}=x+3{{\left( xy \right)}^{\dfrac{1}{3}}}{{x}^{\dfrac{1}{3}}}+3{{\left( xy \right)}^{\dfrac{1}{3}}}{{y}^{\dfrac{1}{3}}}+y$.
Taking factor ${{\left( xy \right)}^{\dfrac{1}{3}}}$common, we get
$\Rightarrow {{\alpha }^{3}}=\left( x+y \right)+3{{\left( xy \right)}^{\dfrac{1}{3}}}\left( {{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}} \right).....(1)$.
Now we will find the value of $xy$,
$\Rightarrow {{\left( xy \right)}^{\dfrac{1}{3}}}={{\left( \left( 45+29\sqrt{2} \right)\times \left( 45-29\sqrt{2} \right) \right)}^{\dfrac{1}{3}}}$.
$\Rightarrow {{\left( xy \right)}^{\dfrac{1}{3}}}={{\left( 2025-1035\sqrt{2}+1035\sqrt{2}-1682 \right)}^{\dfrac{1}{3}}}$.
On simplifying, we get
$\Rightarrow {{\left( xy \right)}^{\dfrac{1}{3}}}={{\left( 343 \right)}^{\dfrac{1}{3}}}$
As we know that,$343={{7}^{3}}$
So, ${{\left( xy \right)}^{\dfrac{1}{3}}}={{\left( {{7}^{3}} \right)}^{\dfrac{1}{3}}}$
On solving we get
$\Rightarrow {{\left( xy \right)}^{\dfrac{1}{3}}}=7$ ---(2).
Let us substitute equation (2) in equation (1).
$\Rightarrow {{\alpha }^{3}}=\left( x+y \right)+3\left( 7 \right)\left( {{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}} \right)$.
But we already know that $\alpha ={{x}^{\dfrac{1}{3}}}+{{y}^{\dfrac{1}{3}}}$such that $x=45+29\sqrt{2}$, $\,y=45-29\sqrt{2}$.
$\Rightarrow {{\alpha }^{3}}=\left( \,\,45+29\sqrt{2}+45-29\sqrt{2} \right)+21\alpha $.
On solving, we get
$\Rightarrow {{\alpha }^{3}}=90+21\alpha $.
$\Rightarrow {{\alpha }^{3}}-21\alpha -90=0$.
∴ We have found the value of ${{\alpha }^{3}}-21\alpha -90$ as 0.

Note: We need to perform each step carefully in order to avoid confusion and calculation mistakes. Whenever we this type of problems, we assign variables for the terms present inside the roots as it will make the simplification easier. We can also solve this problem by using the conjugate surds of one of the terms present in the square root. Similarly, we can expect problems to find the value of ${{\alpha }^{3}}-1$..