Question

. If x² - 6√3 x + 1 = 0 then find the value of x³ + 1/x³

Answer 1

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

Given that

$\rm \: {x}^{2} - 6 \sqrt{3}x + 1 = 0$

$\rm \: {x}^{2} + 1 = 6 \sqrt{3}x$

$\rm \: \dfrac{ {x}^{2} + 1}{x} = 6 \sqrt{3}$

can be further rewritten as

$\rm \: \dfrac{ {x}^{2}}{x} + \dfrac{1}{x} = 6 \sqrt{3}$

$\rm\implies \:x + \dfrac{1}{x} = 6 \sqrt{3}$

On cubing both sides, we get

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

We know,

$\boxed{\tt{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y) \: }}$

So, using this identity, we get

$\rm \: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times x \times \dfrac{1}{x} \times \bigg(x + \dfrac{1}{x} \bigg) = 648 \sqrt{3}$

$\rm \: {x}^{3} + \dfrac{1}{ {x}^{3} } + 3 \times 6 \sqrt{3} = 648 \sqrt{3}$

$\rm \: {x}^{3} + \dfrac{1}{ {x}^{3} } +18 \sqrt{3} = 648 \sqrt{3}$

$\rm \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 648 \sqrt{3} - 18 \sqrt{3}$

$\rm\implies \: \:\boxed{\tt{ \: \rm \: {x}^{3} + \dfrac{1}{ {x}^{3} } = 630 \sqrt{3} \: }}$

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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)