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SSC CHSL 6 Dec 2015 Shift 2 Paper

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Question : 146 of 200

Marks: +1, -0

If

(2a - 1)^2 + (4b - 3)^2

+ (4c + 5)^2 = 0

, Then the value of

\frac{a^3 + b^3 + c^3 - 3abc}{a^2 + b^2 + c^2}

is:

$1\frac{3}{8}$
$3\frac{3}{8}$
$2\frac{3}{8}$
0

Solution:

Given :

(2a - 1)^2 + (4b - 3)^2

+ (4c + 5)^2 = 0

∵ Sum of 3 positive terms is 0, then each term is equal to '0'.

⇒

2a - 1 = 0

⇒

a = \frac{1}{2} = \frac{2}{4}

Similarly,

b = \frac{3}{4}

and

c = -\frac{5}{4}

Now,

(a+b+c) = \frac{2}{4} + \frac{3}{4}

+ \left(-\frac{5}{4}\right) = 0

----(i)

Using,

a^3 + b^3 + c^3 - 3abc

= (a + b + c)

(a^2 + b^2 + c^2 - ab - bc - ac)

⇒

a^3 + b^3 + c^3 - 3abc

(0)(a^2 + b^2 + c^2 - ab - bc - ac)

[Using equation (i)]

⇒

a^3 + b^3 + c^3 - 3abc = 0

--------(ii)

To find :

\frac{a^3 + b^3 + c^3 - 3abc}{a^2 + b^2 + c^2}

=

\frac{0}{a^2 + b^2 + c^2}

= 0 [Using equation (ii)]

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