Question

If 3x + y/5 = 10 and xy = 5 then find 27x3 + y3/125

Answer 1

Correct Question :-

$\sf If\; 3x + \dfrac{y}{5} = 10 \; and \; xy = 5 \; then\;find \; 27x^{3} + \dfrac{y^{3}}{125}$

Solution :-

❒ Here, we will do cubing on both sides of the first equation.  

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$\sf \implies \bigg\{ 3x + \dfrac{y}{5} \bigg\}^{3} = 10^{3}$  

$\bf \maltese \;\; ( a + b )^{3} = a^{3} + b^{3} + 3ab( a + b )$

$\sf \implies \{ 3x \}^{3} + \bigg\{ \dfrac{y}{5} \bigg\}^{3} + \bigg\{ 3 \times 3x \times \dfrac{y}{5} \bigg\} \bigg\{ 3x + \dfrac{y}{5} \bigg\} = 1000$

$\bf \maltese \;\; putting\; 'xy'\; as \;5$ 

$\bf \maltese \;\; putting\; ' \;3x + \dfrac{y}{5} \; ' \;\; as 10$

$\sf \implies 3x^{3} + \dfrac{y^{3}}{125} + \{ 9 \times 10 \} = 1000$

$\sf \implies 3x^{3} + \dfrac{y^{3}}{125} + 90 = 1000$

$\sf \implies 3x^{3} + \dfrac{y^{3}}{125} = 1000 -90$  

$\boxed{\bf{ \bigstar \;\; 3x^{3} + \dfrac{y^{3}}{125} = 910 }}$

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Hence,  

• The answer is 910  

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