Aptitude - Surds and Indices - Discussion

Discussion Forum : Surds and Indices - General Questions (Q.No. 8)

Surds and Indices

1	+	1	+	1	= ?
1 + x^{(b - a)} + x^{(c - a)}		1 + x^{(a - b)} + x^{(c - b)}		1 + x^{(b - c)} + x^{(a - c)}

x^{a - b - c}

None of these

Answer: Option

Explanation:

Given Exp. =

	1 +	x^b	+	x^c
		x^a		x^a

	1 +	x^a	+	x^c
		x^b		x^b

	1 +	x^b	+	x^a
		x^c		x^c

=	x^a	+	x^b	+	x^c
	(x^a + x^b + x^c)		(x^a + x^b + x^c)		(x^a + x^b + x^c)

=	(x^a + x^b + x^c)
	(x^a + x^b + x^c)

= 1.

Discussion:

24 comments Page 2 of 3.

Akshita Tank said: 10 years ago

Can anybody explain the question in easier way?

Achitsa said: 9 years ago

First solve the denominator using LCM the answer you get. Find its reciprocal.

Favour said: 9 years ago

I do not understand any of them.

Franklin Emeka said: 9 years ago

I am not understanding this. Any of them can explain me?

Ekta said: 8 years ago

In the second line,the denominator is solved as:-

(1+x^b/x^a+x^c/x^a) (solving 1st denominator of the first fraction)
Which comes as( x^a+x^b+x^c)/x^a.

Now, the first fraction is : 1/((x^a+x^b+x^c)/x^a) which can be also written as x^a/(x^a+x^b+x^c).

Similarly other two fractions are also solved and finally x^a/(x^a+x^b+x^c) + x^b/(x^a+x^b+x^c) + x^c/(x^a+x^b+x^c) = 1.

Garima said: 8 years ago

Short cut method:
Let a=b=c,
We get , 1/3 +1/3 +1/3 = 3/3 =1.

Garima said: 8 years ago

Short cut method.

Let a=b=c=1.
We get 1/3 +1/3 +1/3 =3/ 3 =1.

(1)

Ayush said: 8 years ago

@Garima

How? Please explain in detail.

Aarambh said: 8 years ago

I cannot understand the second step. Please explain me.

Ragav said: 7 years ago

Here, assign x=1.

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