Question

if x is equal to 3 and Y is equal to -1 find the values of each of the following using identity​

Answer 1

$\mathfrak{\large{\underline{\underline{Answer:-}}}}$

$\boxed{\bf{ \left[ \dfrac{x}{4} - \dfrac{y}{3} \right] \left[ \dfrac{ {x}^{2}}{16} + \dfrac{xy}{12} + \dfrac{ {y}^{2} }{9} \right] = \dfrac{793}{1728} }}$

$\mathfrak{\large{\underline{\underline{Explanation:-}}}}$

$\sf{ \left[ \dfrac{x}{4} - \dfrac{y}{3} \right] \left[ \dfrac{ {x}^{2}}{16} + \dfrac{xy}{12} + \dfrac{ {y}^{2} }{9} \right] }$

It can be written as :

$\sf{= \left[ \dfrac{x}{4} - \dfrac{y}{3} \right] \left[ {( \dfrac{x}{4})}^{2} + \dfrac{x}{4} \times \dfrac{y}{3} + {( \dfrac{y}{4}) }^{2} \right] }$

We know that, a³ - b³ = (a - b)(a² + ab + b²)

Here $\sf{a = \dfrac{x}{4} \: and \: b = \dfrac{y}{3} }$

By substituting the values in the identity we have,

$\sf{= { \left[ \dfrac{x}{4} \right] }^{3} - { \left[ \dfrac{y}{3} \right] }^{3} }$

$\sf{= \dfrac{ {x}^{3} }{64} - \dfrac{ {y}^{3} }{27} }$

If x = 3, y = - 1

$\sf{= \dfrac{ {3}^{3} }{64} - \dfrac{ { (- 1)}^{3} }{27} }$

$\sf{= \dfrac{27}{64} - (\dfrac{ - 1}{27}) }$

$\sf{ = \dfrac{27}{64} + \dfrac{1}{27} }$

Least Common Multiple of 64 and 27 is 1728

$\sf{= \dfrac{27 \times 27}{64 \times 27} + \dfrac{1 \times 64}{27 \times 64} }$

$\sf{ = \dfrac{729}{1728} + \dfrac{64}{1728} }$

$\sf{= \dfrac{729 +64}{1728} }$

$\sf{= \dfrac{793}{1728} }$

$\boxed{\bf{ \left[ \dfrac{x}{4} - \dfrac{y}{3} \right] \left[ \dfrac{ {x}^{2}}{16} + \dfrac{xy}{12} + \dfrac{ {y}^{2} }{9} \right] = \dfrac{793}{1728} }}$