Question

If $a+\dfrac{1}{a}=4;$ find
1). ${{a}^{2}}+\dfrac{1}{{{a}^{2}}}$
2). ${{a}^{4}}+\dfrac{1}{{{a}^{4}}}$
3). ${{a}^{3}}+\dfrac{1}{{{a}^{3}}}$

Answer 1

Hint: For (1) when we squared both sides. ${{\left( a+\dfrac{1}{a} \right)}^{2}}={{\left( 4 \right)}^{2}}\ \Rightarrow \ {{a}^{2}}+\dfrac{1}{{{a}^{2}}}+2.a\dfrac{1}{a}=16$ $\Rightarrow \ {{a}^{2}}+\dfrac{1}{{{a}^{2}}}=14.$
For (2) again, squaring of ${{a}^{2}}+\dfrac{1}{{{a}^{2}}}=14,$ both sides, we get the value of ${{a}^{4}}+\dfrac{1}{{{a}^{4}}}{{.}^{{}}}$
For (3) on cubicying of $a+\dfrac{1}{a}=4,$ then we find ${{a}^{3}}+\dfrac{1}{{{a}^{3}}}.$

Complete step by step solution: Given, $a+\dfrac{1}{a}=4.$

(1). To find :- ${{a}^{2}}+\dfrac{1}{{{a}^{2}}}$
We know that; ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
$\Rightarrow \ \text{on}\ \text{squaring}\ \text{both}\ \text{sides,}\ {{\left( a+\dfrac{1}{a} \right)}^{2}}={{\left( 4 \right)}^{2}}$
$\Rightarrow \ \text{Here,}\ a=a,\ b=\dfrac{1}{a}$
then,
$\Rightarrow \ {{a}^{2}}+\dfrac{1}{{{a}^{2}}}+2.a.\dfrac{1}{a}=16$
$\Rightarrow \ {{a}^{2}}+\dfrac{1}{{{a}^{2}}}=16.2$
$\Rightarrow \ {{a}^{2}}+\dfrac{1}{{{a}^{2}}}=14.$

(2) To find ${{a}^{4}}+\dfrac{1}{{{a}^{4}}}.$
Now,
On squaring of both sides, we get;
${{\left( {{a}^{2}}+\dfrac{1}{{{a}^{2}}} \right)}^{2}}={{\left( 14 \right)}^{2}}$
$\Rightarrow \ {{a}^{4}}+\dfrac{1}{{{a}^{4}}}+2.{{a}^{2}}.\dfrac{1}{{{a}^{2}}}=196$
$\Rightarrow \ {{a}^{4}}+\dfrac{1}{{{a}^{4}}}=196-2$
$\Rightarrow \ {{a}^{4}}+\dfrac{1}{{{a}^{4}}}=194.$

(3) `To find ${{a}^{3}}+\dfrac{1}{{{a}^{3}}}$
On cubicying both sides, we get;
${{\left( a+\dfrac{1}{a} \right)}^{3}}={{\left( 4 \right)}^{3}}$
$\Rightarrow \ \text{we}\ \text{know}\ \text{that;}$
${{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$
So,
$\Rightarrow \ {{a}^{3}}+\dfrac{1}{{{a}^{3}}}+3.a.\dfrac{1}{a}\left( a+\dfrac{1}{a} \right)=64$
$\Rightarrow \ {{a}^{3}}+\dfrac{1}{{{a}^{3}}}+3\left( a+\dfrac{1}{a} \right)=64$
Given, $a+\dfrac{1}{a}=4.$
then,
$\Rightarrow \ {{a}^{3}}+\dfrac{1}{{{a}^{3}}}+3.4=64$
$\Rightarrow \ {{a}^{3}}+\dfrac{1}{{{a}^{3}}}=64-12$
$\Rightarrow \ {{a}^{3}}+\dfrac{1}{{{a}^{3}}}=52$

Note: While solving this type of question, we remembered the ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\ \text{and}\ {{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)$ and use this formula in expression.