If \(x^4+\frac{1}{x^4}=727\), then the value of \(x^3-\frac{1}{x^3}\).

Method II:

Answer: \(140\)

Solution:

We know that if \(x^4+\frac{1}{x^4}=a\), then \(x^2+\frac{1}{x^2}=\sqrt{a+2}\) and \((x-\frac{1}{x})=\sqrt{\sqrt{a+2}-2}\).

Here, we have \(x^4+\frac{1}{x^4}=727\)

Therefore,

\(x^2+\frac{1}{x^2}=\sqrt{727+2}=\sqrt{729}=27\)

\((x-\frac{1}{x})=\sqrt{\sqrt{727+2}-2}\)

\(\Rightarrow (x-\frac{1}{x})=\sqrt{\sqrt{729}-2}\)

\(\Rightarrow (x-\frac{1}{x})=\sqrt{27-2}\)

\(\Rightarrow (x-\frac{1}{x})=\sqrt{25}=5\).

Also, \(x^3-\frac{1}{x^3}=(x-\frac{1}{x})(x^2+\frac{1}{x^2}+1)\).

Therefore,

\(x^3-\frac{1}{x^3}=5 \times (27+1)\)

\(\Rightarrow x^3-\frac{1}{x^3}=5 \times 28\)

\(\Rightarrow  x^3-\frac{1}{x^3}=140\).

Therefore, If \(\mathbf{x^4+\frac{1}{x^4}=727}\), then the value of \(\mathbf{x^3-\frac{1}{x^3}}\) is \(\mathbf{140}\).

Method I:

Answer: \(\mathbf{140}\)

Solution:

Step 1: Use the identity and find the value of \(\mathbf{x^2+\frac{1}{x^2}}\):

Given: \(x^4+\frac{1}{x^4}=727\)

By adding \(2\) both sides we get

\(x^4+\frac{1}{x^4}+2=727+2\)

\(\Rightarrow (x^2)^2+\frac{1}{(x^2)^2}+2\cdot x^2\cdot \frac{1}{x^2}=729\)

\(\Rightarrow (x^2)^2+\left(\frac{1}{x^2}\right)^2+2\cdot x^2\cdot \left(\frac{1}{x}\right)^2=729\)

Here, \(a=x^2, b=\left(\frac{1}{x^2}\right)^2\), then the above equation becomes

\(\Rightarrow a^2+b^2+2ab=729\)

\(\Rightarrow (a+b)^2=729\)

\(\Rightarrow (a+b)^2=27^2\)

\(\Rightarrow (a+b)=27\)              \(\mathbf{[ ∵ \text{If }  a^m= b^n,  \text{then} \, a=b]}\)

As \(a=x^2, b=\left(\frac{1}{(x^2)}\right)^2\), substituting its value back we get

\(\Rightarrow (x^2+\frac{1}{x^2})=27\) .

Step 2: Find the value of  \(\mathbf{ x-\frac{1}{x}}\):

Subtracting \(2\), both sides in above equation, we get

\(\Rightarrow (x^2+\frac{1}{x^2}-2)=27-2\)

\(\Rightarrow (x^2+\left(\frac{1}{x}\right)^2-2\cdot x \cdot \frac{1}{x})=25\)

\(\Rightarrow (x-\frac{1}{x})^2=25\)

\(\Rightarrow (x-\frac{1}{x})^2=5^2\)

\(\Rightarrow (x-\frac{1}{x})=5\).                \(\mathbf{[ ∵ \text{If }  a^m= b^n,  \text{then} \, a=b]}\)

Step 3: Use the identity to find the value of \(\mathbf{ x^3-\frac{1}{x^3}}\):

\( x^3-\frac{1}{x^3}=(x-\frac{1}{x})(x^2+\frac{1}{x^2}+x \cdot \frac{1}{x})\)

\(\mathbf{[∵  (a^3-b^3)=(a-b)(a^2+b^2+ab)]}\)

Substitute the value of  \((x-\frac{1}{x})=5\) and \((x^2+\frac{1}{x^2})=27\) and simplify.

\(\therefore \quad  x^3-\frac{1}{x^3}=(5)(27+1)\)

\(\Rightarrow   x^3-\frac{1}{x^3}=5 \times 28\)

\(\Rightarrow   x^3-\frac{1}{x^3}=140\)

Therefore, If \(\mathbf{x^4+\frac{1}{x^4}=727}\), then the value of \(\mathbf{x^3-\frac{1}{x^3}}\) is \(\mathbf{140}\).