Question

If x is a non-zero real number and x⁴+1/x⁴=194, Find the value of
a) x³+1/x³
b) x³-1/x³​

Answer 1

S O L U T I O N

$\begin{gathered}\sf{( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} - 2 = 194 \\ \\\sf { ({x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = 194 + 2 \\ \\\sf {x}^{2} + \frac{1}{{x}^{2} } = \sqrt{196 } \\ \\ \sf {x}^{2} + \frac{1}{ {x}^{2} } \sf= 14...(1.) \\ \\ \sf using \: {(a + b)}^{2} - 2 = {a}^{2} + {b}^{2} \\ \\\sf ⇝ {( {x} + \frac{1}{ {x} } )}^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\\bf by \: eq(1.) \\ \\ ⇝ {(x + \frac{1}{x} )}^{2} - 2 = 14 \\ \\ ⇝ {(x + \frac{1}{x} )}^{2} = 14 + 2 \\ \\\sf ⇝ x + \frac{1}{x} = \sqrt{16 } \\ \\\sf ⇝ x + \frac{1}{x} = 4..(2.) \\ \\ \\\sf using \: {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3 \times a \times b(a + b) \\ \\\sf So \: as\: per \:the\:Question-:\\ \\ \sf{(x + \frac{1}{x} )}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) \\ \\\sf by \: eq(2.) \\ \\\sf {4}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 \\ \\ \sf 64 - 12 = {x}^{3} + \frac{1}{ {x}^{3} } \\ \\ ⇝ \sf {x}^{3} + \frac{1}{ {x}^{3} } = 52\end{gathered}$

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

Answer 2

Step-by-step explanation:

S O L U T I O N

$\begin{gathered}\sf{( {x}^{2} + \frac{1}{ {x}^{2} } )}^{2} - 2 = 194 \\ \\\sf { ({x}^{2} + \frac{1}{ {x}^{2} } )}^{2} = 194 + 2 \\ \\\sf {x}^{2} + \frac{1}{{x}^{2} } = \sqrt{196 } \\ \\ \sf {x}^{2} + \frac{1}{ {x}^{2} } \sf= 14...(1.) \\ \\ \sf using \: {(a + b)}^{2} - 2 = {a}^{2} + {b}^{2} \\ \\\sf ⇝ {( {x} + \frac{1}{ {x} } )}^{2} - 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\\bf by \: eq(1.) \\ \\ ⇝ {(x + \frac{1}{x} )}^{2} - 2 = 14 \\ \\ ⇝ {(x + \frac{1}{x} )}^{2} = 14 + 2 \\ \\\sf ⇝ x + \frac{1}{x} = \sqrt{16 } \\ \\\sf ⇝ x + \frac{1}{x} = 4..(2.) \\ \\ \\\sf using \: {(a + b)}^{3} = {a}^{3} + {b}^{3} + 3 \times a \times b(a + b) \\ \\\sf So \: as\: per \:the\:Question-:\\ \\ \sf{(x + \frac{1}{x} )}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times x \times \frac{1}{x} (x + \frac{1}{x} ) \\ \\\sf by \: eq(2.) \\ \\\sf {4}^{3} = {x}^{3} + \frac{1}{ {x}^{3} } + 3 \times 4 \\ \\ \sf 64 - 12 = {x}^{3} + \frac{1}{ {x}^{3} } \\ \\ ⇝ \sf {x}^{3} + \frac{1}{ {x}^{3} } = 52\end{gathered}$

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