Question

If x-1/x=√5 find x^3+1/x^3

Answer 1

Answer:

18

Step-by-step explanation:

${\underline{{\text{Given}}}}$

$x - \dfrac{1}{x} = \sqrt{5}$

${\underline{{\text{To Find:}}}}$

${x}^{3} + \dfrac{1}{ {x}^{3} }$

${\underline{{\text{Solution:}}}}$

$x - \frac{1}{x} = \sqrt{5} \: \text{[Squaring both side]} \\ \\ {(x - \frac{1}{x} )}^{2} = {( \sqrt{5}) }^{2} \\ \\ {(x + \frac{1}{x} )}^{2} - 4.x. \frac{1}{x} = 5 \text{ [Using Formula } {(a + b)}^{2} - 4ab = {(a - b)}^{2} ] \\ \\ {(x + \frac{1}{x} )}^{2} - 4. \cancel{x}. \frac{1}{\cancel{x}} = 5 \\ \\ {(x + \frac{1}{x} )}^{2} = 5 + 4 \\ \\ {(x + \frac{1}{x} )}^{2} = 9 \\ \\ {(x + \frac{1}{x} )}^{2} = {(3)}^{2} \\ \\ x + \frac{1}{x} = 3.......(i) \\ \text{[Now, cubing both side ]} \\ {(x + \frac{1}{x}) }^{3} = {( 3)}^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3.x. \frac{1}{x} (x + \frac{1}{x} ) = 27 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3.\cancel{x}. \frac{1}{ \cancel{x}.} (3) = 27 \: \: \: \text{[From (i)]} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } +3.3 = 27 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 27 - 9 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 18 \\ \\ \therefore \text{The required answer is 18 }$

${{\red{\boxed{ \text{\blue{Some Important Formula Of Algebra}}}}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy$

${(x - y)}^{2}={x}^{2}+{y}^{2}-2xy$

${(x + y)}^{2}=(x - y)^{2}+4xy$

${(x-y)}^{2}=(x+y)^{2}-4xy$

$(x + y)^{2}+ (x - y)^{2}=2({x}^{2}+{y}^{2})$

$(x+y)^{2}-(x-y)^{2}=4xy$

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

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