Question

If $(a-\frac{1}{a})=7$; find the value of $(a^2+\frac{1}{a^2})$,$(a^2-\frac{1}{a^2})$ and $(a^3-\frac{1}{a^3})$.

Answer 1

Step 1: Find the value of $(a^2+\frac{1}{a^2})$
Given Expression is $(a-\frac{1}{a})=7$ .....(1)
Now, find the square of $(a-\frac{1}{a})$ by using the formula
$(a-b{)}^2=a^2-2ab+b^2$

$\Rightarrow {(a-\frac{1}{a})}^2=a^2+\frac{1}{a^2}-2\left(a\right)\left(\frac{1}{a}\right)$

$\Rightarrow a^2+\frac{1}{a^2}={(a-\frac{1}{a})}^2+2$

$\Rightarrow a^2+\frac{1}{a^2}=7^2+2$

$\Rightarrow a^2+\frac{1}{a^2}=49+2$

$\therefore a^2+\frac{1}{a^2}=51$ ....(2)

Step 2: Find the value of $(a^2-\frac{1}{a^2})$
Now, find the value of$(a+\frac{1}{a})$ by using the formula $(a+b{)}^2=(a-b{)}^2+4ab$

$\Rightarrow {(a+\frac{1}{a})}^2={(a-\frac{1}{a})}^2+4\left(a\right)\left(\frac{1}{a}\right)$

$\Rightarrow {(a+\frac{1}{a})}^2={(a-\frac{1}{a})}^2+4$

$\Rightarrow {(a+\frac{1}{a})}^2=49+4=53$

$\therefore a+\frac{1}{a}=\pm \sqrt{53}$ .....(3)

Now, multiply equation (1) and (3),

$(a-\frac{1}{a})(a+\frac{1}{a})=\pm 7\sqrt{53}$

$\therefore (a^2-\frac{1}{a^2})=\pm 7\sqrt{53}$

Step 3: Find the value of $(a^3-\frac{1}{a^3})$.
Now, the find the cube of $(a-\frac{1}{a})$ by using the formula
$(a-b{)}^3=a^3-b^3-3ab(a-b)$

$\Rightarrow {(a-\frac{1}{a})}^3=\left(a^3\right)-{\left(\frac{1}{a}\right)}^3-3\left(a\right)\left(\frac{1}{a}\right)(a-\frac{1}{a})$

$\Rightarrow (7{)}^3=a^3-\frac{1}{a^3}-3(7)$

$\therefore a^3-\frac{1}{a^3}=343+21=364$

Hence, the value $(a^2+\frac{1}{a^2})$,$(a^2-\frac{1}{a^2})$ and $(a^3-\frac{1}{a^3})$ are $51,\pm 7\sqrt{53}$ and $364$.