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Question: Prove that the real root of the equation \({{x}^{3}}+12x-12=0\) is \(2\sqrt[3]{2}-\sqrt[3]{4}\) ....

Prove that the real root of the equation x3+12x−12=0{{x}^{3}}+12x-12=0 is 223−432\sqrt[3]{2}-\sqrt[3]{4} .

Explanation

Solution

Hint:Start with the left-hand side of the equation given in the question and put the value of x to be 223−432\sqrt[3]{2}-\sqrt[3]{4} . If the left-hand side gets simplified to zero, i.e., equal to the right-hand side, then 223−432\sqrt[3]{2}-\sqrt[3]{4} is the root of the equation.

Complete step-by-step answer:

The equation x3+12x−12=0{{x}^{3}}+12x-12=0 is a cubic polynomial. For a number to be the root of the root, it must satisfy the polynomial, i.e., if we put the number in place of x in the equation, the left-hand side of the equation is equal to the right-hand side of the equation. So, for 223−432\sqrt[3]{2}-\sqrt[3]{4} to be one of the roots of x3+12x−12=0{{x}^{3}}+12x-12=0 , must satisfy the equation.
∴(223−43)3+12(223−43)−12=0\therefore {{\left( 2\sqrt[3]{2}-\sqrt[3]{4} \right)}^{3}}+12\left( 2\sqrt[3]{2}-\sqrt[3]{4} \right)-12=0
We know (a−b)3=a3−b3−3a2b+3b2a{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3{{a}^{2}}b+3{{b}^{2}}a . So, solving the left-hand side of our equation, we get
(223)3−(43)3−3(223)2(43)+3(43)2(223)+12(223−43)−12{{\left( 2\sqrt[3]{2} \right)}^{3}}-{{\left( \sqrt[3]{4} \right)}^{3}}-3{{\left( 2\sqrt[3]{2} \right)}^{2}}\left( \sqrt[3]{4} \right)+3{{\left( \sqrt[3]{4} \right)}^{2}}\left( 2\sqrt[3]{2} \right)+12\left( 2\sqrt[3]{2}-\sqrt[3]{4} \right)-12
=(223)3−(43)3−3(223)2(43)+3(43)2(223)+12(223−43)−12={{\left( 2\sqrt[3]{2} \right)}^{3}}-{{\left( \sqrt[3]{4} \right)}^{3}}-3{{\left( 2\sqrt[3]{2} \right)}^{2}}\left( \sqrt[3]{4} \right)+3{{\left( \sqrt[3]{4} \right)}^{2}}\left( 2\sqrt[3]{2} \right)+12\left( 2\sqrt[3]{2}-\sqrt[3]{4} \right)-12
=16−4−12(163)+6(323)+2423−1243−12=16-4-12\left( \sqrt[3]{16} \right)+6\left( \sqrt[3]{32} \right)+24\sqrt[3]{2}-12\sqrt[3]{4}-12
=−24(23)+12(43)+2423−1243=-24\left( \sqrt[3]{2} \right)+12\left( \sqrt[3]{4} \right)+24\sqrt[3]{2}-12\sqrt[3]{4}
= 0
Therefore, we can say that 223−432\sqrt[3]{2}-\sqrt[3]{4} is a root of the equation x3+12x−12=0{{x}^{3}}+12x-12=0.

Note: A root can graphically be defined as the values of x for which the curve of the equation cuts the x-axis, i.e., the value of y is zero. It is prescribed to go with the method of polynomial division when you are dealing with a polynomial with higher degrees as it might be difficult to put the value of x and see whether the equation is satisfied or not. In some questions, the fact that complex roots appear in pairs provided the coefficients of the polynomial are real may also be very helpful.