Question

if x^2+y^2=4 and x-y=3 then x^3-y^3=?​

Answer 1

Answer :-

Value of x³ - y³ is 9/2.

Explanation :-

Finding the value of xy

We know that

(x - y)² = x² + y² - 2xy

Here

• x - y = 3

• x² + y² = 4

By substituting the values

⇒ 3² = 4 - 2xy

⇒ 9 = 4 - 2xy

⇒ 2xy = 4 - 9

⇒ 2xy = - 5

⇒ xy = - 5/2

Finding the value of x³ - y³

We know that

x³ - y³ = (x - y)(x² + y² + xy)

Here

• x - y = 3

• x² + y² = 4

• xy = - 5/2

By substituting the values

⇒ x³ - y³ = 3{ 4 + (-5/2)}

⇒ x³ - y³ = 3(4 - 5/2)

⇒ x³ - y³ = 3{(8 - 5)/2}

⇒ x³ - y³ = 3(3/2)

⇒ x³ - y³ = 9/2

∴ the value of x³ - y³ is 9/2.

Answer 2

SOLUTION:-

Given:

•x² + y² =4

•x - y =3

•x³ - y³ =?

Therefore,

Here,

Squaring both sides;

=) (x-y)² = (3)²

=) x² + y² - 2xy= 9

=) 4 - 2xy= 9 [given:x² +y²= 4]

=) 4- 9= 2xy

=) -5= 2xy

=) xy= -5/2

Again,

Cubic both sides;

=) (x-y)³= (3)³

=) x³ - y³ - 3xy(x-y)= 27

$= > {x}^{3} - {y}^{3} - 3( - \frac{5}{2} )(3) = 27 \\ \\ = > {x}^{3} - {y}^{3} - ( - \frac{15}{2} )(3) = 27 \\ \\ = > {x}^{3} - {y}^{3} + \frac{45}{2} = 27 \\ \\ = > {x}^{3} - {y}^{3} = 27 - \frac{45}{2} \\ \\ = > {x}^{3} - {y}^{3} = \frac{54 - 45}{2} \\ \\ = > {x}^{3} - {y}^{3} = \frac{9}{2}$

Thus,

The value of x³ - y³= 9/2.

Hope it helps ☺️