Question

if x⁴+1/x⁴ = 194 find the value of x³+1/x³

Answer 1

Answer:

⇒ x³ + 1/x³ = 52

Step-by-step explanation:

Given: x⁴ + (1/x⁴) = 194

⇒ x⁴ + (1/x⁴) = 196 - 2

⇒ x⁴ + (1/x⁴) + 2 = 196

⇒ (x² + 1/x²)² = 196

⇒ (x² + 1/x²) = 14

It can be written as,

⇒ x² + 1/x² = 16 - 2

⇒ x² + 1/x² + 2 = 16

⇒ (x + 1/x) = 4

On cubing both sides, we get

⇒ (x + 1/x) = (4)³

⇒ x³ + 1/x³ + 3(x + 1/x) = 64

⇒ x³ + 1/x³ + 3(4) = 64

⇒ x³ + 1/x³ + 12 = 64

⇒ x³ + 1/x³ = 64 - 12

⇒ x³ + 1/x³ = 52.

Hope it helps!

Answer 2

Your question :
if x⁴+1/x⁴ = 194 find the value of x³+1/x³

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${( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} - 2 = 194 \\ \\ { ({x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = 194 + 2 \\ \\ {x}^{2} + \frac{1}{{x}^{2} } = \sqrt{196 } \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } = 14........(1.) \\ \\ using \: {(a + b)}^{2} - 2 = {a}^{2} + {b}^{2} \\ \\ = > {( {x} + \frac{1}{ {x} } )}^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ by \: eq(1.) \\ \\ = > {(x + \frac{1}{x} )}^{2} - 2 = 14 \\ \\ = > {(x + \frac{1}{x} )}^{2} = 14 + 2 \\ \\ = > x + \frac{1}{x} = \sqrt{16 } \\ \\ = > x + \frac{1}{x} = 4........(2.) \\ \\ \\ using \: {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3 \times a \times b(a + b) \\ \\ so \\ \\ {(x + \frac{1}{x} )}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) \\ \\ by \: eq(2.) \\ \\ {4}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 \\ \\ 64 - 12 = {x}^{3} + \frac{1}{ {x}^{3} } \\ \\ = > > {x}^{3} + \frac{1}{ {x}^{3} } = 52$

So, Your answer is $\bold{\underline{52}}$

HOPE IT HELPS