Question

plzz solve both for exam ok

Answer 1

Check it in NCERT
u will definitely get your answer

Answer 2

1/

Solution:

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Given:

3a + 2b = 5c,.

&

abc = 0
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To Find :

Value of 27a³ + 8b³ - 125c³

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27a³ + 8b³ - 125c³

=>  (3a)³ + (2b)³ + (-5c)³

we know that,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² -ab -bc -ac)

=> (3a)³ + (2b)³ + (-5c)³ - 3(0) = (3a + 2b - 5c)((3a)² + (2b)² +(-5c)² -(3a)(2b) -(2b)(-5c) - (3a)(-5c))

=> (3a)³ + (2b)³ + (-5c)³ = (0)(9a² + 4b² + 25c² -6ab +10bc +15ac)

=> 27a³ + 8b³ -125c³ = 0

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2/

Solution:

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Given:

$x^2 + \frac{1}{x^2} = 66$

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To Find :

The value of $x^3 - \frac{1}{x^3}$

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We know that,

(a + b)² = a² + 2ab + b²,.

(a+ b)² - 2ab = a² + b²,..

so,

$(x)^2 + ( \frac{1}{x})^2 = 66$ ... (i)

Adding 2 both the sides,

=> $x^2 + \frac{1}{x^2} + 2 = 66 + 2$

=> $x^2 + \frac{1}{x^2} + 2(x)( \frac{1}{x^2} ) = 68$

=> $(x + \frac{1}{x} )^2 = 68$...(ii)

=> $x + \frac{1}{x} = \sqrt{68}$

=> $\frac{x^2 + 1}{x} = \sqrt{68}$

=> $x^2 + 1 = \sqrt{68} x$

=> $x^2 - \sqrt{68} x + 1 = 0$

=> [tex] x^{2} - \sqrt{68}x = -1 [/tex]]

=> $x^2 - 2( \frac{ \sqrt{68}x}{2}) = -1$ (multiplying and dividing by 2)

=> $x^2 - 2( \frac{ \sqrt{68}x}{2}) + (\frac{ \sqrt{68}}{2})^2 = -1+ (\frac{ \sqrt{68}}{2})^2$ (Adding √68/2 both the sides)

=> $(x - \frac{ \sqrt{68}}{2})^2 = -1 + \frac{68}{4}$

=> $(x - \frac{ \sqrt{17(4)}}{2})^2 = \frac{-4+68}{4}$

=> $(x - \frac{ 2\sqrt{17}}{2})^2 = \frac{64}{4}$

=> $(x - \sqrt{17} )^2 = 16$

=> $x - \sqrt{17} = 4$

=> $x = 4 + \sqrt{17}$

And we know that,

a³ - b³ = (a-b)(a² + ab + b²),.

=> $x^3 - \frac{1}{x^3} = (x - \frac{1}{x})(x^2 - (x)( \frac{1}{x})+ ( \frac{1}{x^2}))$

=> $(4+ \sqrt{17} - \frac{1}{4 + \sqrt{17}})(66 + 1)$

=> $(4+ \sqrt{17} - \frac{4- \sqrt{17}}{16-17})(67)$

=> $(4+ \sqrt{17} + 4 - \sqrt{17})(67)$

=> (8)(67)

=> 536

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                                     Hope it Helps!!