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Home » In Print We Trust: These 10 Brands Prove Why Physical Media Still Wins! - Crosslake

In Print We Trust: These 10 Brands Prove Why Physical Media Still Wins! - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Given that $ a + b + c = 0 $, substitute into the identity:
a^3 + b^3 + c^3 - 3abc = 0 \cdot (a^2 + b^2 + c^2 - ab - bc - ca) = 0.
a^3 + b^3 + c^3 = 3abc.
This completes the proof.
oxed{a^3 + b^3 + c^3 = 3abc}
Question: Compute $ \sum_{k=1}^{50} rac{1}{k(k+2)} $.
Solution: We evaluate the sum:
\sum_{k=1}^{50} rac{1}{k(k+2)}.
We use partial fractions:
rac{1}{k(k+2)} = rac{A}{k} + rac{B}{k+2}.

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Match:
\(x^2\): \(b - a + 1 = -2 \Rightarrow b - 4 + 1 = -2 \Rightarrow b = 1...
\(x\): \(a - b + A = 1 \Rightarrow 4 - 1 + A = 1 \Rightarrow A = -2\)
constant: \(b + B = 5 \Rightarrow 1 + B = 5 \Rightarrow B = 4\)
So remainder is \(-2x + 4\)? But earlier substitution gave \(3\omega +...
Wait â recalculate at \(\omega\):
$\omega^2 = \omega - 1$
$\omega^3 = \omega \cdot \omega^2 = \omega(\omega - 1) = \omega^2 - \o...
$x^4 = -\omega$
$3x^3 = 3(-1) = -3$

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