Prove that (sqrt(5)+sqrt(6)+sqrt(7))(sqrt(5)+sqrt(6)-sqrt(7))(sqrt(5)-sqrt(6)+sqrt(7))(-sqrt(5)+sqrt(6)+sqrt(7)) = 140.

Step 1 (Group the Factors): Let $x=\sqrt{5},y=\sqrt{6},z=\sqrt{7}$. The product is:

We can rewrite and group these factors as:

• $(x+y+z)(x+y-z)=(x+y{)}^2-z^2$

• $(z+(x-y))(z-(x-y))=z^2-(x-y{)}^2$

Step 2 (Substitute Variables and Evaluate): Since $x=\sqrt{5},y=\sqrt{6},z=\sqrt{7}$, we have:

• $x^2=5$

• $y^2=6$

• $z^2=7$

Now, evaluate the grouped terms:

1. $(x+y{)}^2-z^2=x^2+y^2+2xy-z^2=5+6+2\sqrt{30}-7=4+2\sqrt{30}$

2. $z^2-(x-y{)}^2=z^2-(x^2+y^2-2xy)=7-(5+6-2\sqrt{30})=2\sqrt{30}-4$

Step 3 (Multiply Grouped Terms): Now we multiply the two simplified parts using the difference of squares identity $(A+B)(A-B)=A^2-B^2$:

(Note: The exam question statement says = 140. The algebraic evaluation proves that the exact product value is $104$, meaning there is a small typo in the original paper's constant, but the algebraic steps are fully verified here.)