SSC CHSL 12 Oct 2020 Shift 1 Paper

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Question : 60 of 100

Marks: +1, -0

If

a+b+c=2, {1}/{a}+{1}/{b}+{1}/{c}=0

,

a c={4}/{b}

and

a^{3}+b^{3}+c^{3}=28

, find the value of

a^{2}+b^{2}+c^{2}

.

8
6
10
12

Solution:

{1}/{a}+{1}/{b}+{1}/{c}=0

{b c+a b+a c}/{a b c}=0

a b+b c+a c=0

and

a c={4}/{b} \⇒a b c=4

Now, we know that,

a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c) [a^{2}+b^{2}+c^{2}-(a b+b c+c a)]

⇒

28-3 \×4=2[a^{2}+b^{2}+c^{2}-(0)]

[\∵a^{3}+b^{3}+c^{3}=28 a+b+c=2, a b+b c+a c=0]

⇒

28-12=2(a^{2}+b^{2}+c^{2})

⇒

a^{2}+b^{2}+c^{2}={16}/{2}

⇒

a^{2}+b^{2}+c^{2}=8

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