Question

If x is not equals to 0 and x+1/x=2; then show that:-
x^2+1/x^2=x^3+1/x^3=x^4+1/x^4​

Answer 1

Required Answer:-

Question:

• If x ≠ 0 and x + 1/x = 2, prove that x² + 1/x² = x³ + 1/x³ = x⁴ + 1/x⁴

Proof:

Given that,

➡ x + 1/x = 2

➡ (x² + 1)/x = 2

➡ x² + 1 = 2x

➡ x² - 2x + 1 = 0

➡ (x)² - 2 × (x) × (1) + (1)² = 0

➡ (x - 1)² = 0

On solving, we get x = 1

Therefore, x² + 1/x²

= (1)² + 1/(1)²

= 2

Similarly,

x³ + 1/x³ = 2 and x⁴ + 1/x⁴ = 2

Therefore,

➡ x² + 1/x² = x³ + 1/x³ = x⁴ + 1/x⁴ = 2 (Proved)

Answer 2

${ \huge \color{red}{\underline{ \underline {\rm \color{black}{Given}}}} : }$

$\large\sf \: x + \frac{1}{x} = 2$

${\large {\underline{ \underline{ \rm {To \: show \: that }}} : }}$

$\sf {x}^{2} + \frac{1}{ {x}^{2} } = {x}^{3} + \frac{1}{ {x}^{3} } = {x}^{4} + \frac{1}{ {x}^{4} }$

${ \huge \color{blue}{ \underline {\underline{ \rm \color{black}{Solution}}}}:}$

Squaring above equation's both sides

$\sf(x + \frac{1}{x} {)}^{2} = ( {2})^{2} ....(i)$

We know the identity=$\sf{(a + b {)}^{2} = {a}^{2} + {b}^{2} + 2(ab)}$

$\rightarrow \sf ({x})^{2} + ( \frac{1}{x} {)}^{2} + 2(x. \frac{1}{x} ) = ( {2})^{2}$

$\sf \rightarrow {x}^{2} + \frac{1}{ {x}^{2} } + 2 \cancel{x}. \frac{1}{ \cancel{ x} } = 4$

$\sf \rightarrow {x}^{2} + \frac{1}{ {x}^{2} } = 4 - 2 \\ \\ \implies \sf {x}^{2} + \frac{1}{ {x}^{2} } = 2....(ii)$

Now, cubing both the sides

$\sf(x + \frac{1}{x} {)}^{3} = ({2})^{3}$

We know the identity=$\sf(a + b {)}^{3} = ({a})^{3} + ({b})^{2} + 3(ab)(a + b)$

$\sf \rightarrow( {x})^{3} + ( \frac{1}{x} {)}^{3} + 3 \cancel x. \frac{1}{ \cancel x} (x + \frac{1}{x} ) = ( {2})^{3}$

$\rightarrow \sf {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 2 = 8$

$\rightarrow\sf{x}^{3}+\frac{1}{x}^{3}=2$

Now squaring (i) and (ii)

$\rightarrow\sf({x}^{2}+\frac{1}{x}^{2})={2}^{2}$

$\sf\rightarrow{{x}^{2}}^{2}+\frac{{(1)}^{2}}{{x}^{2}}+2\cancel{x}^{2}.\frac{1}{x}^{2}=4$

$\sf\rightarrow {x}^{4}+\frac{1}{{x}^{4}}=4-2$

$\rightarrow\sf{x}^{4}+\frac{1}{{x}^{4}}=2$

#Hence Proved