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∣ ∣xCrxCr+1xCr+2yCryCr+1yCr+2zCrzCr+1zCr+2∣ ∣=∣ ∣ ∣xCrx+1Cr+1x+2Cr+2yCry+1Cr+1y+2Cr+2zCrz+1Cr+1z+2Cr+2∣ ∣ ∣

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Solution

Hint: First observe that
xCr+xCr+1=x+1Cr+1
and xCr+1+xCr+2=x+1Cr+2.
Now applying C3+C2 and C2+C1, we get
Δ=∣ ∣ ∣xCrx+1Cr+1x+1Cr+2yCry+1Cr+1y+1Cr+2zCrz+1Cr+1z+1Cr+2∣ ∣ ∣
Again applying C3+C2, we get
Δ=∣ ∣ ∣xCrx+1Cr+1x+2Cr+2yCry+1Cr+1y+2Cr+2zCrz+1Cr+1z+2Cr+2∣ ∣ ∣
by the same rule.

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