Question : If a + b + c = 19, ab + bc + ca = 120, then what is the value of a3 + b3 + c3 – 3abc?
Option 1: 18
Option 2: 23
Option 3: 31
Option 4: 19
Correct Answer: 19
Solution : Given, a + b + c = 19 and ab + bc + ca = 120 We know, a3 + b3 + c3 – 3abc = (a + b + c) [(a + b + c)2 – 3(ab + bc + ca)] = 19[192 – 3 × 120] = 19[361 – 360] = 19 × 1 = 19 Hence, the correct answer is 19.
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Question : If a + b + c = 1, ab + bc + ca = –1, and abc = –1, then what is the value of a3 + b3 + c3?
Option 1: 1
Option 2: 5
Option 3: 3
Option 4: 2
Question : If a + b + c = 5 and ab + bc + ca = 7, then the value of a3 + b3 + c3 – 3abc is:
Option 1: 15
Option 2: 20
Option 3: 25
Option 4: 30
Question : If a + b + c = 6 and a2 + b2 + c2 = 38, then what is the value of a(b2 + c2) + b(c2 + a2) + c(a2 + b2) + 3abc?
Option 1: 3
Option 2: 6
Option 3: –6
Option 4: –3
Question : If a + b + c = 10 and a2 + b2 + c2 = 48, then the value of ab + bc + ca is _______.
Option 1: 25
Option 2: 26
Option 3: 24
Option 4: 18
Question : If $a+b+c=1, ab+bc+ca=-1$ and $abc=-1$, then the value of $a^{3}+b^{3}+c^{3}$ is:
Option 2: – 1
Option 3: 2
Option 4: – 2
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