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Algebraic Identities

If x-y=7 an...

Question

If $x - y = 7 a n d x^{3} - y^{3} = 133$ ; find:

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

A

29

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B

44

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C

76

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D

13

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Solution

The correct option is A 29
Let,the two number be $x$ and $y$
Therefore,we get
$x - y = 7$
$x^{3} - y^{3} = 133$
Now,
$x - y = 7$
Squaring both sides,
$=> (x - y)^{2} = 7^{2}$
$=> x^{2} + y^{2} - 2 x y = 49$
$=> x^{2} + y^{2} = 49 + 2 x y$
Now,
$x^{3} - y^{3} = 133$
$=> (x - y) (x^{2} + y^{2} + x y) = 133$
$=> 7 (49 + 2 x y + x y) = 133$
$=> 49 + 3 x y = \frac{133}{7}$
$=> 3 x y = 19 - 49$
$=> x y = \frac{- 30}{3}$
$=> x y = - 10$
Putting the value of $x y$ we get,
$x^{2} + y^{2} = 49 + 2 (- 10)$
$=> x^{2} + y^{2} = 49 - 20$
$=> x^{2} + y^{2} = 29$

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Similar questions

x - y = 7 a n d x^{3} - y^{3} = 133

x y

Q. Find the value of

x^{3} + y^{3} + z^{3} - 3 x y z

x + y + z = 12

x^{2} + y^{2} + z^{2} = 70

u = {tan}^{- 1} (\frac{x^{3} + y^{3}}{x - y})

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = ?

(x^{2} + y^{2} = 58)

(x + y = 10)

, then find the value of

(x^{3} + y^{3})

{(x + y)}^{3} - z^{2} = 4

{(y + z)}^{2} - x^{2} = 9

{(z + x)}^{2} - y^{2} = 36

, then find the value of

x + y + z

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