To solve the problem, we need to find the value of \((2 + \sqrt{5})^4 - (2 - \sqrt{5})^4\).
- Let's apply the formula for the difference of powers:
a^4 - b^4 = (a^2 + b^2)(a^2 - b^2)
- Here, \(a = (2 + \sqrt{5})\) and \(b = (2 - \sqrt{5})\).
- First, find \(a^2\) and \(b^2\):
- \((2 + \sqrt{5})^2 = 2^2 + 2 \times 2 \times \sqrt{5} + (\sqrt{5})^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5}\)
- \((2 - \sqrt{5})^2 = 2^2 - 2 \times 2 \times \sqrt{5} + (\sqrt{5})^2 = 4 - 4\sqrt{5} + 5 = 9 - 4\sqrt{5}\)
- Calculate \(a^2 + b^2\):
(9 + 4\sqrt{5}) + (9 - 4\sqrt{5}) = 9 + 9 = 18
- Calculate \(a^2 - b^2\):
(9 + 4\sqrt{5}) - (9 - 4\sqrt{5}) = 9 + 4\sqrt{5} - 9 + 4\sqrt{5} = 8\sqrt{5}
- Substitute these values back into the formula:
a^4 - b^4 = (a^2 + b^2)(a^2 - b^2) = 18 \times 8\sqrt{5} = 144\sqrt{5}
Thus, the value of \((2 + \sqrt{5})^4 - (2 - \sqrt{5})^4\) is \(\mathbf{144\sqrt{5}}\), which matches the correct option.
Similar Questions
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If $a + b + c = 0$, then find the value of $\frac{(a^2+b^2+c^2)^2}{a^2b^2+b^2c^2+c^2a^2}$
- If $x^4 + \frac{1}{x^4} = 322$, then $x^3 - \frac{1}{x^3} = ?$
- What is the value of the following expression:
$(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)(1 - x)$ - If $a + b + c = 17$, $abc = 168$, and $ab + bc + ca = 94$, then $a^3 + b^3 + c^3 = ?$
- If $a = \frac{b^2}{b - a}$, then the value of $a^3 + b^3$ is:
- If $ab = 10$ and $a^{2} + b^{2} = 29$, then $(a - b)^{2} = ?$
- If $(x + \frac{1}{x}) = 7$, then $(x - \frac{1}{x})$ is equal to:
- Find the value of $\frac{(74 + 47)^2 + (74 - 47)^2}{74^2 + 47^2}$
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Important Questions from Identities
The coefficient of y in the expansion of (2y – 5) 3, is:
If x + y = 2 and \(\frac{1}{x}+\frac{1}{y}=\frac{18}{5}\) , then the value of (x 3+ y 3) is:
If x - y = 11 and \(\rm \frac{1}{x} - \frac{1}{y} = \frac{11}{24}\) then the value of x 3 - y 3 + x 2y 2 ?
If 2x 2- 8x - 1 = 0, then what is the value of \(\rm 8x^3 - \frac{1}{x^3}\) ?
If \(\rm x+ \frac{1}{x} = 4,\) then the value of \(\rm x^5 + \frac{1}{x^5}\) is:
