If a+b = 3 and a-b = 1, find the value of \(a^4 - b^4\).
15
We are given \(a + b = 3\) and \(a - b = 1\), and need \(a^4 - b^4\).
Factor using the difference of squares twice: \(a^4 - b^4 = (a^2)^2 - (b^2)^2 = (a^2 + b^2)(a^2 - b^2)\).
Factor the second bracket again: \(a^2 - b^2 = (a + b)(a - b)\), so \(a^4 - b^4 = (a^2 + b^2)(a + b)(a - b)\).
Solve the linear pair by adding and subtracting: adding gives \(2a = 4 \Rightarrow a = 2\), and then \(b = 1\).
So \(a^2 + b^2 = 2^2 + 1^2 = 4 + 1 = 5\).
Substitute all parts: \(a^4 - b^4 = 5 \times 3 \times 1 = 15\).
Factoring into \((a^2 + b^2)(a + b)(a - b)\) lets us use the given sums directly. The value of \(a^4 - b^4\) is 15.
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