Question

If \[{a^2} + \dfrac{1}{{{a^2}}} = 47\], find \[a + \dfrac{1}{a},\]\[{a^3} + \dfrac{1}{{{a^3}}}\]

Answer 1

Hint: If the left-hand side of an equation is equal to the right side for any values of the variable, then that equation is an identity which is used in algebraic expansions and factorizations. In standard identities, the values are equal to each other regardless of what values are substituted for the variables.
Algebraic equations which hold for all values of the variable in them are called algebraic identities. They are also used for the factorization of polynomials.
In this question to find the value of\[a + \dfrac{1}{a}\], use the form\[{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy\]; here equate the identities and find the vales.
Also to find \[{a^3} + \dfrac{1}{{{a^3}}}\], use the form \[{\left( {x + y} \right)^3} = {x^3} + {y^3} + 3xy\left( {x + y} \right)\]and substitute in the equation.

Complete step-by-step answer:
\[{a^2} + \dfrac{1}{{{a^2}}} = 47\]
One of the algebraic identities says\[{\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy\]
 Where, \[x = a\] and \[y = \dfrac{1}{a}\]
Now substitute the value of x & y equation, we get
\[
  {\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy \\
  {\left( {a + \dfrac{1}{a}} \right)^2} = {a^2} + {\left( {\dfrac{1}{a}} \right)^2} + 2\left( {a \times \dfrac{1}{a}} \right) \\
  {\left( {a + \dfrac{1}{a}} \right)^2} = {a^2} + \dfrac{1}{{{a^2}}} + 2 \\
 \]
Since, \[{a^2} + \dfrac{1}{{{a^2}}} = 47\]
Hence we can write:
\[
  {\left( {a + \dfrac{1}{a}} \right)^2} = {a^2} + \dfrac{1}{{{a^2}}} + 2 \\
  {\left( {a + \dfrac{1}{a}} \right)^2} = 47 + 2 \\
  a + \dfrac{1}{a} = \sqrt {49} \\
  a + \dfrac{1}{a} = 7 \\
 \]
Therefore, the value of \[\left( {a + \dfrac{1}{a}} \right)\] is 7.
Now for\[{a^3} + \dfrac{1}{{{a^3}}}\]
Using another algebraic identity is given as: \[{\left( {x + y} \right)^3} = {x^3} + {y^3} + 3xy\left( {x + y} \right)\]
By comparing \[\left( {a + \dfrac{1}{a}} \right)\] with $ \left( {x + y} \right) $ it can be written as:
\[x = a\] and \[y = \dfrac{1}{a}\]
Now substitute the value in the algebraic equation \[{\left( {x + y} \right)^3} = {x^3} + {y^3} + 3xy\left( {x + y} \right)\]
\[
  {\left( {x + y} \right)^3} = {x^3} + {y^3} + 3xy\left( {x + y} \right) \\
  {\left( {a + \dfrac{1}{a}} \right)^3} = {a^3} + {\left( {\dfrac{1}{a}} \right)^3} + 3 \times \left( {a \times \dfrac{1}{a}} \right)\left( {a + \dfrac{1}{a}} \right) \\
  {\left( {a + \dfrac{1}{a}} \right)^3} = {a^3} + {\left( {\dfrac{1}{a}} \right)^3} + 3\left( {a + \dfrac{1}{a}} \right) \\
 \]
As we have already got the value of \[a + \dfrac{1}{a} = 7\]
Hence, by substitution, it can be written as
\[
  {\left( {a + \dfrac{1}{a}} \right)^3} = {a^3} + {\left( {\dfrac{1}{a}} \right)^3} + 3\left( {a + \dfrac{1}{a}} \right) \\
  {\left( 7 \right)^3} = {a^3} + \dfrac{1}{{{a^3}}} + 3\left( 7 \right) \\
  343 = {a^3} + \dfrac{1}{{{a^3}}} + 21 \\
  {a^3} + \dfrac{1}{{{a^3}}} = 343 - 21 \\
  {a^3} + \dfrac{1}{{{a^3}}} = 322 \\
 \]
Therefore, the value of\[{a^3} + \dfrac{1}{{{a^3}}} = 322\]

Note: Students can check whether a given equation is the identity or not by transforming either one side of an equation in such a way that they become equal to the other side of the equation. Some of the algebraic identities in maths are:
Identity 1: \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
Identity 2: \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\]
Identity 3: \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\]
Identity 4: \[\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab\]