TS ICET 23 May 2018 Shift 1 Paper

Question : 77 of 200

Marks: +1, -0

If a + b + c = 0 then

x^{a^{2}b^{-1}c^{-1}}

x^{a^{-1}b^{2}c^{-1}}

x^{a^{-1}b^{-1}c^{2}}

Solution:

a+b+c=0⇒ a^{3}+b^{3}+c^{3}=3abc

Now,

x {a^{2}}/{b c} × x{b^{2}}/{a c} × x {c^{2}}/{a b}=x{a^{3}}/{a b c} × x {b^{3}}/{a b c} × x {c^{3}}/{a b c}

=x^{{(a^{3}+b^{3}+c^{3})}/{a b c}}=x^{{3 a b c}/{a b c}}=x^{3}

Go to Question:

Prev Question

Next Question