Question

if $a-\frac{1}{a}=2$, then prove that
${a}^{8} - 34 {a}^{4} + 1 = 0$

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Answer 1

Given -

• $\sf a-\frac{1}{a}=2$

To prove -

• $\sf a^8 - 34a^4 + 1=0$

Proof -

• $\sf a-\frac{1}{a}=2$

Squaring both sides

• Apply Identity
• (a - b)² = a² - 2ab + b²

$\\\implies\sf a^2 - 2\times{a}\times\dfrac{1}{a} +\dfrac{1}{a^2} = 4$

$\implies\sf a^2 - 2 + \dfrac{1}{a^2}=4$

$\implies\sf a^2 + \dfrac{1}{a^2}=4+2$

$\implies\sf a^2 + \dfrac{1}{a^2}=6$

Squaring both sides again

• Apply Identity
• (a + b)² = a² + 2ab + b²

$\implies\sf a^4 + 2\times{a^2}\times\dfrac{1}{a^2}+\dfrac{1}{a^4}=36$

$\implies\sf a^4 + 2 +\dfrac{1}{a^4}=36$

$\implies\sf a^4 +\dfrac{1}{a^4}=36-2$

$\implies\sf a^4 +\dfrac{1}{a^4}=34$

$\implies\sf a^4 +\dfrac{1}{a^4}-34=0$

$\implies\sf \dfrac{a^8 +1 - 34a^4}{a^4}=0$

• aᵐ × aⁿ = aᵐ⁺ⁿ

$\\\\\implies\sf a^8 +1 - 34a^4=0$

$\implies\sf a^8 -34a^4+1 = 0$

• $\large{\underline{\boxed{\tt{\therefore a^8-34a^4+1=0\:\: Proved}}}}$

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Answer 2

Answer:

Given :-

$\sf \: a - \dfrac{1}{a}$

To Proof :-

${a}^{8} - 34 {a}^{4} + 1 = 0$

Required Proof

• Identity = (a - b)² = a² - 2ab + b²

$\huge \sf \: \: {a}^{2} - 2a \times \dfrac{1}{a} \times \dfrac{1}{ {a}^{2} } = 4$

$\huge\sf \: {a}^{2} - 2 + \dfrac{1}{a {}^{2} } = 4$

$\huge\sf \: {a}^{2} + \dfrac{1}{ {a}^{2} } = 4 + 2$

$\huge \sf \: {a}^{2} + \frac{1}{ {a}^{2} } = 6$

Let's

Again square both sides by using identity

(a + b)² = a² + 2ab + b²

$\huge \sf \: {a}^{4} + 2 {a}^{2} \times \frac{1}{ {a}^{2} } + \frac{1}{ {a}^{4} } = 36$

$\huge \sf \: {a}^{4} + 2 + \frac{1}{ {a}^{4} } = 36$

$\huge \sf \: {a}^{4} + \frac{1}{ {a}^{4} } = 36 - 2$

$\huge \sf \: {a}^{4} + \frac{1}{ {a}^{4} } = 34$

$\huge \sf \: {a}^{4} + \frac{1}{ {a}^{4} } - 34 = 0$

$\huge \sf \: \frac{ {a}^{8} + 1 - 34 {a}^{4} }{a {}^{4} } = 0$

$\huge \sf \: {a}^{8} + 1 - 34 {a}^{4} = 0$

$\huge \sf \: {a}^{8} - 34 {a}^{4} + 1 = 0$

Hence proved.