Question

Class 9 ,Algebra ,Expansions​

Answer 1

GIVEN :-

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

TO FIND :-

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

SOLUTION:-

We know ,

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

Then,

$x + \dfrac{1}{x} = \sqrt{3}$

We shall do cubing on both sides

$(x + \dfrac{1}{x} ) {}^{3} = (\sqrt{3} ) {}^{3}$

(a + b)³ = a³ + b³ + 3ab(a + b)

$(x + \dfrac{1}{x} ) {}^{3} = x {}^{3} + \dfrac{1}{x {}^{3} } + 3 \times x \times \dfrac{1}{x} (x + \dfrac{1}{x} )$

$(x + \dfrac{1}{x} )^3= x {}^{3} + \dfrac{1}{x {}^{3} } + 3(x + \dfrac{1}{x} )$

$( \sqrt{3} ) {}^{3} = x {}^{3} + \dfrac{1}{x {}^{3} } + 3 \sqrt{3}$

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

$x {}^{3} + \dfrac{1}{x {}^{3} } = 0$

KNOW MORE :-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc