Question

answer this easy question and earn 49 points

Answer 1

$<b>Answer :</b>$

➡️ x^2 + 1/x^2 = 7
➡️ x^3 + 1/x^3 = 18
➡️ x^4 + 1/x^4 =47
$<b>Step-by-step explanation :</b>$

x+1/x=3

squaring both sides

x2+1/x2=32

x2+1/x2=9

x2+1/x2=9-2

x2+1/x2=7

_____________

To find :- x^3 + 1/x^3

Given: x+1/x = 3

Cubing both sides

(x+1/x)^3 = 3^3

Now, using the formula ( a+b)^3

x^3 + 1/x^3 + 3× x × 1/x ( x+ 1/x) = 27

x^3 + 1/x^3 + 3 (x+1/x) = 27

A/q: x+ 1/x = 3

x^3 + 1/x^3 + 3 (3) = 27

x^3 + 1/x^3 = 27-9

x^3 + 1/x^3 = 18

______________
➡️ x^4 + 1/x^4

➡️ Given

x+1x=3x+1x=3

Taking square on both sides

(x+1x)2=32x+1x)2=32

x2+1x2+2=9x2+1x2+2=9

x2+1x2=9−2x2+1x2=9−2

x2+1x2=7x2+1x2=7

Again taking square on both sides, we have

(x2+1x2)2=72(x2+1x2)2=72

(x2)2+(1x2)2+2=49(x2)2+(1x2)2+2=49

x4+1x4=49−2x4+1x4=49−2

x4+1x4=47

____________

Hope it will help you

@thanksforquestion

@bebrainly

Answer 2

$\textbf{\huge{ANSWER:}}$

$\sf{Given:}$

${x} + \frac{1}{ {x}} = 3$

Now,

${(x + \frac{1}{x} )}^{2} = {3}^{2} \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 =9 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 7$

Next,

${(x + \frac{1}{x} )}^{3} = {3}^{3} \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 3(x + \frac{1}{x} ) = 27 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } + 9 = 27 \\ \\ = > {x}^{3} + \frac{1}{ {x}^{3} } = 18$

At last,

${( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = {7}^{2} \\ \\ = > {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 49 \\ \\ = > {x}^{4} + \frac{1}{ {x}^{4} } = 47$
Identities used:
${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

Hope it Helps!!