Question

if X +1/7=7 and xy=84 find the value of x^3+1​

Answer 1

Answer:

$x + \frac{1}{x} = 7 \\ xy = 84 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } \\ = (x + \frac{1}{x} )( {x}^{2} - x \times \frac{1}{x} + \frac{1}{ {x}^{2} } ) \\ = (x + \frac{1}{x} )( {x}^{2} + \frac{1}{ {x}^{2} } - 1) \\ = (x + \frac{1}{x} )( {x + \frac{1}{x} )}^{2} - 1 - 1 \\ = 7( {7 }^{2} - 2) \\ = 7 \times (49 - 2) \\ = 7 \times 47 \\ = 329$

Answer 2

322

Step-by-step explanation:

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

$x + \dfrac{1}{x} = 7 \\ \\ xy = 84 \: \: \text{[We don't need it]}$

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

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

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

$x + \frac{1}{x} = 7......(1) \: \: [\text{Cubing both side} ] \\ \\ {(x + \frac{1}{x} )}^{3} = {7}^{3} \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3.x. \frac{1}{x} (x + \frac{1}{x} ) = 343 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3 .\cancel{x} . \frac{1}{\cancel{x}} (7) = 343 \: \: [\text{From (1)} ] \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 3.7 = 343 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } + 21 = 343 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 343 - 21 \\ \\ {x}^{3} + \frac{1}{ {x}^{3} } = 322$

Hence, the required answer is 322.

${{\boxed{ \text{\blue{Some Important Formula}}}}}$

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \\ \\ {(x + y)}^{2} = (x - y) {}^{2} + 4xy \\ \\ {(x - y)}^{2} = (x + y) {}^{2} - 4xy \\ \\ (x + y) {}^{2} + (x - y) {}^{2} = 2( {x}^{2} + {y}^{2} ) \\ \\ (x + y) {}^{2} - (x - y) {}^{2} = 4xy \\ \\ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \\ \\ (x - y) {}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y)$