Question

determine the value of x2+y2 when x3-y3=54,x-y=18,xy=2​

Answer 1

Question:

Determine the value of  x² + y²  when x³ - y³ = 54,  x - y = 18,  xy = 2​.

Solution:

We may recall the identity given below.

$\Largex^3-y^3=(x-y)(x^2+xy+y^2)$

So,

    x³ - y³ = 54

⇒  (x - y)(x² + xy + y²) = 54

⇒  18 (x² + xy + y²) = 54

⇒  x² + xy + y² = 54 / 18

⇒  x² + xy + y² = 3

⇒  x² + y² + 2 = 3

⇒  x² + y² = 3 - 2

⇒  x² + y² = 1

Hence, 1 is the answer.

Answer 2

Answer:

$\boxed{\textbf{\large{x square + y square = 1}}}$

Given:

${x}^{3} - {y}^{3} = 54$

$x - y = 18$

$xy = 2$

To find :

${x}^{2} + {y}^{2}$

Explanation:

☑ we know the identity

${a}^{3} - {b}^{3} = (a - b) \\ ( {a}^{2} + ab + {b}^{2} )$

☑ here, for x and y

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

☑ put the given values,

$= > 54 = (18)( {x}^{2} + (2) + </p><p>{y}^{2} )$

$= > \frac{54}{18} = {x}^{2} + {y}^{2} + (2)$

$= > 3 - 2 = {x}^{2} + {y}^{2}$

➡ 1 = x^2 + y^2

$\boxed{\textbf{\large{x square + y square = 1}}}$