Question

If x>0 and x² +1/9x² = 25/36, find x³+ 1/27x³​

Answer 1

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{9 {x}^{2} } = \dfrac{25}{36}$

$\rm :\longmapsto\: {x}^{2} + \dfrac{1}{ {(3x)}^{2} } = \dfrac{25}{36}$

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{3x} \bigg) }^{2} - 2 \times x \times \dfrac{1}{3x} = \dfrac{25}{36}$

$\green{\boxed{ \bf{ \: \because \: {x}^{2} + {y}^{2} = {(x + y)}^{2} - 2xy}}}$

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{3x} \bigg) }^{2} - \dfrac{2}{3} = \dfrac{25}{36}$

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{3x} \bigg) }^{2} = \dfrac{25}{36} + \dfrac{2}{3}$

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{3x} \bigg) }^{2} = \dfrac{25 + 24}{36}$

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{3x} \bigg) }^{2} = \dfrac{49}{36}$

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{3x} \bigg) }^{2} = \dfrac{ {7}^{2} }{ {6}^{2} }$

$\rm :\implies\:x + \dfrac{1}{3x} = \dfrac{7}{6} \: \: \: as \: x \: > \: 0$

On cubing both sides, we get

$\rm :\longmapsto\: {\bigg(x + \dfrac{1}{3x} \bigg) }^{3} = {\bigg(\dfrac{7}{6} \bigg) }^{3}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {27x}^{3} } + 3 \times x \times \dfrac{1}{3x}\bigg(x + \dfrac{1}{3x} \bigg) = \dfrac{343}{216}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {27x}^{3} } + \dfrac{7}{6} = \dfrac{343}{216}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {27x}^{3} } = \dfrac{343}{216} - \dfrac{7}{6}$

$\rm :\longmapsto\: {x}^{3} + \dfrac{1}{ {27x}^{3} } = \dfrac{343 - 252}{216}$

$\bf :\longmapsto\: {x}^{3} + \dfrac{1}{ {27x}^{3} } = \dfrac{91}{216}$

More Identities to know:

• (a + b)² = a² + 2ab + b²

• (a - b)² = a² - 2ab + b²

• a² - b² = (a + b)(a - b)

• (a + b)² = (a - b)² + 4ab

• (a - b)² = (a + b)² - 4ab

• (a + b)² + (a - b)² = 2(a² + b²)

• (a + b)³ = a³ + b³ + 3ab(a + b)

• (a - b)³ = a³ - b³ - 3ab(a - b)