Question-226118

Question Number 226118 by Spillover last updated on 20/Nov/25

Answered by mr W last updated on 20/Nov/25

$$\mathrm{1}+\alpha+\beta+\gamma+\alpha\beta+\beta\gamma+\gamma\alpha+\alpha\beta\gamma=\mathrm{33} \\ $$$$\mathrm{1}+\mathrm{6}+\alpha\beta+\beta\gamma+\gamma\alpha+\alpha\beta\gamma=\mathrm{33} \\ $$$$\alpha\beta+\beta\gamma+\gamma\alpha+\alpha\beta\gamma=\mathrm{26}\:\:\:…\left({i}\right) \\ $$$$\left(\alpha+\beta+\gamma\right)^{\mathrm{3}} =\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} +\gamma^{\mathrm{2}} +\mathrm{3}\left(\alpha+\beta+\gamma\right)\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)−\mathrm{3}\alpha\beta\gamma \\ $$$$\mathrm{6}^{\mathrm{3}} =\mathrm{87}+\mathrm{18}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)−\mathrm{3}\alpha\beta\gamma \\ $$$$\mathrm{6}\left(\alpha\beta+\beta\gamma+\gamma\alpha\right)−\alpha\beta\gamma=\mathrm{43}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\alpha\beta+\beta\gamma+\gamma\alpha=\frac{\mathrm{69}}{\mathrm{7}} \\ $$$$\Rightarrow\alpha\beta\gamma=\frac{\mathrm{113}}{\mathrm{7}} \\ $$$$\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}+\frac{\mathrm{1}}{\gamma}=\frac{\alpha\beta+\beta\gamma+\gamma\alpha}{\alpha\beta\gamma}=\frac{\mathrm{69}}{\mathrm{7}}×\frac{\mathrm{7}}{\mathrm{113}}=\frac{\mathrm{69}}{\mathrm{113}}=\frac{{m}}{{n}} \\ $$$$\Rightarrow{m}+{n}=\mathrm{69}+\mathrm{113}=\mathrm{182}\:\checkmark \\ $$

Answered by Spillover last updated on 20/Nov/25

$$ \\ $$*Solution demystified* 🔥

Given: A + B + C = 6 —- (1)
A³ + B³ + C³ = 87 —- (2)
(A + 1)(B + 1)(C + 1) = 33 — (3)
Where A, B, and C represent Alpha, Beta and Gamma, respectively

(A + B + C)³ = [(A + B) + C]³
= (A + B)³ + 3C(A + B)² + 3C²(A + B) + C³ = (A³ + 3A²B + 3AB² + B³) + 3C(A² + B² + 2AB) + 3C²(A + B) + C³
= (A³ + B³ + C³) + 3(A + B + C)(AB + BC + AC) – 3ABC

Let A + B + C = p, AB + BC + AC = q and ABC = r

=> r³ = (A³ + B³ + C³) + 3pq – 3r — (4)

Substituting values in eqn (4), we have that:
6³ = 87 + 3•6q – 3r

=> 18q – 3r = 129
=> 6q – r = 43 —- (5)

From eqn (3), (A + 1)(B + 1)(C + 1) = 33
= ABC + AB + AC + A + BC + B + C + 1
= ABC + (AB + BC + AC) + (A + B + C) + 1 = r + q + p + 1
= r + q + 6 + 1

=> r + q + 7 = 33
=> q + r= 26 —— (6)

Add eqn (5) and eqn (6), it follows that:
7q = 69, => q = 69/7

Substitute q in eqn (6),
r = 26 – q = 26 – 69/7 = 113/7
r = 113/7 🦦

Now, 1/A + 1/B + 1/C
= (BC + AC + AB)/(ABC)
= q/r = (69/7)/(113/7) = 69/113

*: . 1/A + 1/B + 1/C = 69/113*

*By comparison, m = 69 & n = 113, => m + n = 69 + 113 = 182*

*: . m + n = 182*. 🦦✨

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