Question

a-1/a=3

find a^2+3a+1/a^2-3a

Answer 1

Answer:

squaring both sides

a2 + 1/a2-2=9

a2+1/a2=9+2

a2+1/a2=11

Therefore 3a gets cancelled by 3a

answer is 11

Answer 2

Given

$\sf \to \bigg(a - \dfrac{1}{a} \bigg) = 3$

To find the value of

$\sf \to \: {a}^{2} + 3a + \dfrac{1}{ {a}^{2} } - 3a$

$\sf \to \: {a}^{2} + \cancel{3a} + \dfrac{1}{ {a}^{2} } - \cancel{ 3a}$

$\sf \to \: {a}^{2} + \dfrac{1}{ {a}^{2} }$

So We have to find the value of

$\sf \to \: {a}^{2} + \dfrac{1}{ {a}^{2} }$

So Now take

$\sf \to \bigg(a - \dfrac{1}{a} \bigg) = 3$

Squaring on both side

$\sf \to \bigg(a - \dfrac{1}{a} \bigg) ^{2} = (3) {}^{2}$

Use this identities

$\sf \to(a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab$

Now

$\sf \to {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 \times a \times \dfrac{1}{a} = 9$

$\sf \to {a}^{2} + \dfrac{1}{ {a}^{2} } - 2 = 9$

$\sf \to {a}^{2} + \dfrac{1}{ {a}^{2} } = 9 + 2$

$\sf \to {a}^{2} + \dfrac{1}{ {a}^{2} } = 11$

Answer

$\sf \to {a}^{2} + \dfrac{1}{ {a}^{2} } = 11$