fukirua wrote:

Bunuel wrote:

\((x+2)(y+2)(z+2)-xyz=2(xy + xz + yz + 2x + 2y + 2z + 4)\).

The question itself is very clear and not hard but it took me ages to multiple three brackets without a mistake. Is there a trick?

Thanks

Start with multiplying the first two in the product of the three factors:

\((x+2)(y+2)(z+2)=(xy+2x+2y+4)(z+2)\)

No need to fully carry out the other multiplication. You can see that there will be one term of \(xyz\), which will cancel out in the final expression with \(-xyz\).

In addition, you will have terms containing products of two factors - like \(xy, \,\,xz\), and \(yz\). All will have a coefficient of 2 in front.

Then the terms with \(x, \,\,y,\) and \(z\), all have a coefficient of 4 in front.

Answers A, B, and C can be eliminated. Between D and E, obviously E wins.

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