Question-164077

Question Number 164077 by mathlove last updated on 13/Jan/22

Answered by cortano1 last updated on 13/Jan/22

(1) x + y + x y + 1 = 4 \Rightarrow (x + 1) (y + 1) = 4

(2) y + z + y z + 1 = 6 \Rightarrow (y + 1) (z + 1) = 6

(3) z + x + z x + 1 = 8 \Rightarrow (z + 1) (x + 1) = 8

(1) \times (2) \times (3) \Rightarrow (x + 1) (y + 1) (z + 1) = \pm \sqrt{4 \times 4 \times 4 \times 3} = \pm 8 \sqrt{3}

z + 1 = \pm 2 \sqrt{3}

x + 1 = \pm \frac{4 \sqrt{3}}{3}

y + 1 = \pm \sqrt{3}

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