# (solved)Question 1 SSC-CGL 2020 March 4 Shift 3

If x + y + z = 3, and x² + y² + z² = 101, then √ (x³ + y³ + z³ - 3xyz) =?
(Rev. 18-Jun-2024)

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### Question 1SSC-CGL 2020 Mar 4 Shift 3

If x + y + z = 3, and x² + y² + z² = 101, then √ (x³ + y³ + z³ - 3xyz) =?

### Solution by Trick

Observe that there are 3 variables but 2 equations. If we set the third variable to any convenient value, the other two variables will adjust accordingly, and still the 2 given equations will remain satisfied! Let us take z = 0.

We need to find = √ (x³ + y³)

We know x³ + y³ = (x + y)³ - 3xy(x + y)

We already have x + y = 3, x² + y² = 101

but we need xy also.

Now, (x + y)² = x² + y² + 2xy, whence xy = -46

Finally, √(x³ + y³) = √[(3)³ - 3(-46)(3)] = √441 = 21 ans!

### Solution in Detail

This is a formula based question. We know x³ + y³ + z³ - 3xyz = (x+y+z) [(x + y + z)² - 3(xy + yz + zx)]. But we do not have the value of (xy + yz + zx). For this let us use the identity: (x + y + z)² = x² + y² + z² + 2(xy + yz + zx). Substituting the available values, 3² = 101 + 2(xy + yz + zx), which gives (xy + yz + zx) = -46. Finally, we get (x³ + y³ + z³ - 3xyz) = (x+y+z) [(x + y + z)² - 3(xy + yz + zx)] i.e., = (3) [(3)² - 3(-46)] = 441, and the required square root = 21 ans!