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Maths_Expert_3
In order to solve the problem: If \(x^4+\frac{1}{x^4}=727\), then the value of \(x^3-\frac{1}{x^3}\). We need to use algebraic identities...
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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 \(\mathbf{140}\).
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 \(\mathbf{140}\).
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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 \(\mathbf{140}\).
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 \(\mathbf{140}\).
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