Algebra and Applications Test 2

Question : 18 of 40

Marks: +1, -0

If x = 997, y = 998 and z = 999, then the value of

x^ 2 + y ^2 + z ^2 - xy - yz - zx

will be:

Solution:

x = 997, y = 998, z = 999

x^2 + y^2 + z^2 - xy - yz - zx

= (997)^2 + (998)^2 + (999)^2 - 997 × 998

- 998 ×999 - 999 × 997

= 9,940,09 + 9,96,004 + 9,98,001 - 9,95,006 - 9,97,002 - 9,96,003
= 29,88,014 - 29,88,011 = 3

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