Question

plzz solve ques 6 plzzzz

Answer 1

GOOD MORNING!!

( x³ + y³ )= ( x + y )³ - 3xy ( x + y )

=>

( a + 1/a )² = 3

Take square root on both sides we get

( a + 1/a ) = ±√3

Cubing on both sides, we have

( a + 1/a )³ = ( √3 )³

( a³ + 1/a³ ) - 3 ( a + 1/a ) = ( √3 )³

( a³ + 1/a³ ) = ( √3 )³ ±3 ( √3 )

=>

( a³ + 1/a³ ) = ( √3 )³ +3 ( √3 )

OR

( a³ + 1/a³ ) = (√3)³ - 3√3

=>

( a³ + 1/a³ ) = 3√3 + 3√3

Becoz ( √3 )³ = 3√3

OR

( a³ + 1/a³ ) = +3√3 - 3√3

Becoz ( √3)³ = 3√3

So,

Either

( a³ + 1/a³ ) = 6√3

OR

( a³ + 1/a³ ) =0

Answer 2

Step-by-step explanation:

Given :

To prove that, if $(a + \frac{1}{a})^2= 3$,

Then, $a^3 +(\frac{1}{a})^3 = 0$

Solution :

If $(a + \frac{1}{a})^2= 3$,

Then,

$a + \frac{1}{a}=$±$\sqrt{3}$ ..(i)

_

We know that,

a³ + b³ = (a + b) (a² - ab + b²)

$(a)^3 + (\frac{1}{a})^3$

⇒ $(a + \frac{1}{a})(a^2 - (a)(\frac{1}{a} + \frac{1}{a^2})$

⇒ $($±$\sqrt{3})(a^2 - 1 + \frac{1}{a^2})$ ..(ii)

_

$a + \frac{1}{a})^2= 3$

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

$a^2 + \frac{1}{a^2} + 2= 3$

$a^2 + \frac{1}{a^2} + 2 - 3 = 0$

$a^2 + \frac{1}{a^2} -1 = 0$ ..(iii)

_

Hence,

$(a)^3 + (\frac{1}{a})^3$ = $(\sqrt{3})(a^2 - 1 + \frac{1}{a^2})$ = $(\sqrt{3})(0)$ = 0

Hence, proved