Question

Find the value of \[\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\] is

Answer 1

Hint: Put the expression as x. By using the identity \[{{\left( a+b \right)}^{3}}\] expands the form. Solve the polynomial obtained and find the value of x.

Complete step-by-step answer:

Given us the expression, \[\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\].
Let us put, \[x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\].
\[\therefore x=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}-(1)\]
We know the basic formula, \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\].

Let us take a cube on both sides of equation (1).
\[\begin{align}
  & {{x}^{3}}={{\left[ {{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}}}+{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}}} \right]}^{3}} \\
 & {{x}^{3}}={{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}\times 3}}+{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}\times 3}}+3\left[ {{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}}}.{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}}} \right]\left[ {{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}}}+{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}}} \right] \\
 & {{x}^{3}}=\left( 5+2\sqrt{13} \right)+\left( 5-2\sqrt{13} \right)+3\left[ {{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}}}.{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}}} \right]x \\
 & \because x={{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}}}+{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}}} \\
\end{align}\]

The exponent product rule can be used to solve the expression. According to exponent product rule, when multiplied by 2 powers that have some base, you can add the exponent.
i.e. \[{{x}^{m}}.{{x}^{n}}={{x}^{m+n}}\]

But the expression \[\left[ {{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}}}.{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}}} \right]\] doesn’t have same base but they have same power. Thus we can write them as,
\[\begin{align}
  & {{\left( 5+2\sqrt{13} \right)}^{\dfrac{1}{3}}}.{{\left( 5-2\sqrt{13} \right)}^{\dfrac{1}{3}}}={{\left[ \left( 5+2\sqrt{13} \right).\left( 5-2\sqrt{13} \right) \right]}^{\dfrac{1}{3}}} \\
 & \therefore {{x}^{3}}=\left( 5+2\sqrt{13} \right)+\left( 5-2\sqrt{13} \right)+3{{\left[ \left( 5+2\sqrt{13} \right).\left( 5-2\sqrt{13} \right) \right]}^{\dfrac{1}{3}}}x \\
\end{align}\]
\[{{x}^{3}}=10+3{{\left[ {{\left( 5 \right)}^{2}}-{{\left( 2\sqrt{13} \right)}^{2}} \right]}^{\dfrac{1}{3}}}x\] \[\left\{ \because \left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}} \right\}\]
      \[\begin{align}
  & =10+3{{\left[ 25-52 \right]}^{\dfrac{1}{3}}}x \\
 & =10+3{{\left[ -27 \right]}^{\dfrac{1}{3}}}x=10-3{{\left( 27 \right)}^{\dfrac{1}{3}}}x \\
 & =10-3\times 3x=10-9x \\
\end{align}\]
\[\begin{align}
  & \therefore {{x}^{3}}=10-9x \\
 & {{x}^{3}}+9x-10=0 \\
\end{align}\]

Now, we need to get the value of the expression, \[{{x}^{3}}+9x-10=0\].
Let us factorize the polynomial.
Thus we can write, \[{{x}^{3}}+9x-10=0\]as,
\[\left( x-1 \right)\left( {{x}^{2}}+x-10 \right)=0\]
\[\therefore x-1=0\] and \[{{x}^{2}}+x-10=0\].
From x – 1 = 0, we get x = 1.
\[{{x}^{2}}+x-10=0\] is in the form of a quadratic equation, \[a{{x}^{2}}+bx+c=0\].
\[\therefore \] a = 1, b = 1, c = -10.
\[\begin{align}
  & \therefore x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-1\pm \sqrt{1-4\times 1\times -10}}{2} \\
 & x=\dfrac{-1\pm \sqrt{1+40}}{2} \\
\end{align}\]
\[x=\dfrac{-1\pm \sqrt{41}}{2}\]
Thus we can take the value of x as 1.
\[\therefore \] The value of the expression, \[\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}=1\].

Note: The absolute value of the root \[\dfrac{-1\pm \sqrt{41}}{2}\] is greater than 2. On the other hand \[\sqrt[3]{5+2\sqrt{13}}\] is less than 2. The other term \[\sqrt[3]{5-2\sqrt{13}}\] is negative and of smaller absolute value. So we can conclude that 0 < x < 2, and the only possible solution is x = 1.