Question

Solve: \[\left( {{x^{\dfrac{1}{3}}} + {x^{\dfrac{{ - 1}}{3}}}} \right)\left( {{x^{\dfrac{2}{3}}} - 1 + {x^{\dfrac{{ - 2}}{3}}}} \right)\]

Answer 1

Hint:To solve the given problem it is advisable that we must know the basic properties of multiplication, division, subtraction and addition of algebraic equations with unknown variables. We use simple mathematical calculations to solve the given problem. We take two variables $a$ and $b$ as \[{x^{\dfrac{1}{3}}}\] and \[{x^{\dfrac{{ - 1}}{3}}}\] respectively and reduce the given polynomial to get the solution.

Complete step by step answer:
We are given with polynomial \[\left( {{x^{\dfrac{1}{3}}} + {x^{\dfrac{{ - 1}}{3}}}} \right)\left( {{x^{\dfrac{2}{3}}} - 1 + {x^{\dfrac{{ - 2}}{3}}}} \right)\]. Now we write 1 as \[\left( {{x^{\dfrac{1}{3}}}} \right)\left( {{x^{\dfrac{{ - 1}}{3}}}} \right)\] , as we know from rules of multiplication \[\left( {{x^m}} \right)\left( {{x^n}} \right) = {x^{m + n}}\]. Here we take \[m = \dfrac{1}{3}\] and \[n = \dfrac{{ - 1}}{3}\] using simple mathematics \[m + n = 0\].

Now \[{x^{m + n}}\]\[ = {x^0} = 1\]. Therefore the given equation becomes as follows
\[\left( {{x^{\dfrac{1}{3}}} + {x^{\dfrac{{ - 1}}{3}}}} \right)\left( {{x^{\dfrac{2}{3}}} - 1 + {x^{\dfrac{{ - 2}}{3}}}} \right) = \left( {{x^{\dfrac{1}{3}}} + {x^{\dfrac{{ - 1}}{3}}}} \right)\left[ {{{\left( {{x^{\dfrac{1}{3}}}} \right)}^2} - \left( {{x^{\dfrac{1}{3}}}} \right)\left( {{x^{\dfrac{{ - 1}}{3}}}} \right) + {{\left( {{x^{\dfrac{{ - 1}}{3}}}} \right)}^2}} \right]\]
Now let us take two variables a and b and let
\[{x^{\dfrac{1}{3}}} = a\] and \[{x^{\dfrac{{ - 1}}{3}}} = b\] .
On substituting these the given polynomial becomes
\[\left( {{x^{\dfrac{1}{3}}} + {x^{\dfrac{{ - 1}}{3}}}} \right)\left( {{x^{\dfrac{2}{3}}} - 1 + {x^{\dfrac{{ - 2}}{3}}}} \right)\]\[ = (a + b)({a^2} - ab + {b^2})\]

Using simple mathematics we can solve the reduced polynomial in a and b as
\[(a + b)({a^2} - ab + {b^2})\]\[ = {a^3} - {a^2}b + a{b^2} + b{a^2} - a{b^2} + {b^3}\]
\[\Rightarrow (a + b)({a^2} - ab + {b^2}) = {a^3} + {b^3}\]
Now taking a and b values in the final reduced form of the polynomial we get as follows
\[{a^3} + {b^3}\]\[ = {\left( {{x^{\dfrac{1}{3}}}} \right)^3} + {\left( {{x^{\dfrac{{ - 1}}{3}}}} \right)^3}\]
\[\Rightarrow {a^3} + {b^3} = {x^{\dfrac{3}{3}}} + {x^{\dfrac{{ - 3}}{3}}} \\
\Rightarrow {a^3} + {b^3} = x + {x^{ - 1}}\]
\[\therefore {a^3} + {b^3} = x + \dfrac{1}{x}\]

Therefore the solution will be \[\left( {{x^{\dfrac{1}{3}}} + {x^{\dfrac{{ - 1}}{3}}}} \right)\left( {{x^{\dfrac{2}{3}}} - 1 + {x^{\dfrac{{ - 2}}{3}}}} \right)\]\[ = x + \dfrac{1}{x}\].

Additional information: There are several types of equations: polynomial equations, trigonometric equations, logarithmic equations, exponential equations, differential equations and then a mixture of equations consisting of each of them. These help us to represent many real life examples in terms of mathematical expressions and solve accordingly as per our requirements.

Note:The possibility of mistake can be not applying appropriate formulae for calculations. Be careful with the signs and calculations as in such questions, the possibility of making a mistake is either of the sign or a calculation error. We must have a strong grip over the concepts of algebraic expressions and simplification rules.