Question

If then \[{a^2} + 2 = {3^{\dfrac{2}{3}}} + {3^{ - \dfrac{2}{3}}}\] show \[3{a^3} + 9a = 8\]?

Answer 1

Hint: Here in the given equation \[{a^2} + 2 = {3^{\dfrac{2}{3}}} + {3^{ - \dfrac{2}{3}}}\], considering both LHS and RHS terms we can take the common terms and rewrite the equation in simplified manner and find the value of variable a from the equation, then substitute the value of a in \[3{a^3} + 9a\] then we can prove the equation \[3{a^3} + 9a = 8\].

Complete step-by-step solution:
To prove the above equations we must understand the relation between the powers of the sum and difference in which there are certain rules to be followed.
Note that:
\[{\left( {{3^{\dfrac{1}{3}}} - {3^{ - \dfrac{1}{3}}}} \right)^2} = {\left( {{3^{\dfrac{1}{3}}}} \right)^2} - 2\left( {{3^{\dfrac{1}{3}}}} \right)\left( {{3^{ - \dfrac{1}{3}}} + {{\left( {{3^{ - \dfrac{1}{3}}}} \right)}^2}} \right)\]
\[{\left( {{3^{\dfrac{1}{3}}} - {3^{ - \dfrac{1}{3}}}} \right)^2} = {3^{\dfrac{2}{3}}} - 2 + {3^{ - \dfrac{2}{3}}}\]
Let us write the given equation:
\[{a^2} + 2 = {3^{\dfrac{2}{3}}} + {3^{ - \dfrac{2}{3}}}\]
Hence, we can simplify by subtracting 2 from both sides of the equation as
\[{a^2} + 2 - 2 = {3^{\dfrac{2}{3}}} + {3^{ - \dfrac{2}{3}}} - 2\]
Hence, subtracting the terms we get:
\[{a^2} = {3^{\dfrac{2}{3}}} + {3^{ - \dfrac{2}{3}}} - 2\]
\[{a^2} = {\left( {{3^{\dfrac{1}{3}}} + {3^{ - \dfrac{1}{3}}}} \right)^2}\]
Hence, we can write the value of a as:
\[a = \pm \left( {{3^{\dfrac{1}{3}}} + {3^{ - \dfrac{1}{3}}}} \right)\]
If,
\[a = {3^{\dfrac{1}{3}}} + {3^{ - \dfrac{1}{3}}}\] ………………. 1
Then, now let us consider \[3{a^3} + 9a\]
\[3{a^3} + 9a = 3a\left( {{a^2} + 3} \right)\] ………………….. 2
Substitute the value of a from equation 1 in equation 2 we get
\[3{a^3} + 9a = 3\left( {{3^{\dfrac{1}{3}}} - {3^{ - \dfrac{1}{3}}}} \right)\left( {{{\left( {{3^{\dfrac{1}{3}}} - {3^{ - \dfrac{1}{3}}}} \right)}^2} + 3} \right)\] …………. 3
Simplify the terms, by applying formulas in equation 3
= \[3\left( {{3^{\dfrac{1}{3}}} - {3^{ - \dfrac{1}{3}}}} \right)\left( {{3^{\dfrac{2}{3}}} + 1 + {3^{ - \dfrac{2}{3}}}} \right)\]
Multiplying the terms, we get
= \[3\left( {{3^{\dfrac{1}{3}}}\left( {{3^{\dfrac{2}{3}}} + 1 + {3^{ - \dfrac{2}{3}}}} \right) - {3^{ - \dfrac{1}{3}}}\left( {{3^{\dfrac{2}{3}}} + 1 + {3^{ - \dfrac{2}{3}}}} \right)} \right)\]
= \[3\left( {\left( {3 + {3^{\dfrac{1}{3}}} + {3^{ - \dfrac{1}{3}}}} \right) - \left( {{3^{\dfrac{1}{3}}} + {3^{ - \dfrac{1}{3}}} + \dfrac{1}{3}} \right)} \right)\]
= \[3\left( {3 - \dfrac{1}{3}} \right)\]
= \[9 - 1\]
= 8

Therefore, hence proved \[3{a^3} + 9a = 8\]

Note: The key point to solve this type of equation is, when we want to solve an equation including one unknown variable, as in the given example above, we always aim at isolating the unknown variable. It is always a good idea to first isolate the terms including the variable from the constants to begin with as we did above by subtracting or adding before dividing or multiplying away the coefficient in front of the variable. If your equation contains like terms it is preferable to begin by combining the like terms before continuing solving the equation.