Surds and Indices play a major role quantitative aptitude test. There is no competitive exam without the questions from this topic. We have already learned this topic in our lower classes.

Even though we have been already taught this topic in our lower classes, we need to learn some more short cuts which are being used to solve the problems in the above topic.

The only thing we have to do is, we need to apply the appropriate short cut and solve the problems in a limited time. This limited time will be one minute or less than one minute in most of the competitive exams.

To do problems on surd and indices, we have to remembers the following laws on surds and indices.

Students who are preparing to improve their aptitude skills and those who are preparing for this type of competitive test must prepare this topic in order to have better score. Because, today there is no competitive exam without questions from this topic.

Whether a person is going to write placement exam to get placed or a students is going to write a competitive exam in order to get admission in university, they must be prepared to solve problems on indices and surds. This is the reason for why people must study this topic.

As we mentioned in the above paragraph, a person who wants to get placed in a company and a students who wants to get admission in university for higher studies must write competitive exams like placement test and entrance exam. To meet the above requirement, it is very important to score more marks in the above mentioned competitive exams.

To score more marks, they have to prepare the topics like this . Preparing this topic would definitely improve their marks in the above exams. Preparing this topic is not difficult task. We are just going to remember the stuff that we have already learned in our lower classes

Students have to learn few basic operations in this topic and some additional tricks. Already we are much clear with the four basic operations which we often use in math. They are addition, subtraction, multiplication and division. Even though we are much clear with these four basic operations, we have to be knowing some more stuff to do the problems which are being asked from this topic in competitive exams. The stuff which I have mentioned above is nothing but the tricks and shortcuts which need to solve the problems in a very short time.

Here, we are going to have some problems on surds and indices. You can check your answer online and see step by step solution.

1. The value of xp + q + r is

Let x^{1/p} = y^{1/q} = z^{1/r} = k

Then x^{1/p} = k, y^{1/q} = k and z^{1/r} = k

x = k^{p}, y = k^{q} and z = k^{r}

Given: xyz = 1 ---> k^{p}.k^{q}.k^{r} = 1

k^{p + q + r} = 1 ---(1)

We know that a^{0} = 1. So, k^{0} = 1

In (1), plug 1 = k^{0}

k^{p + q + r} = k^{0}

Therefore p+q+r = 0

Then x

x = k

Given: xyz = 1 ---> k

k

We know that a

In (1), plug 1 = k

k

Therefore p+q+r = 0

2. The value of [1-{1-(1-x

[1-{1-(1-x^{2})^{-1}}^{-1}]^{-1/2} = [1-{1-1/(1-x^{2})}^{-1}]^{-1/2}

= [1-{(1-x^{2}-1)/(1-x^{2})}^{-1}]^{-1/2}

= [1-{-x^{2}/(1-x^{2})}^{-1}]^{-1/2}

= [1-{x^{2}/(x^{2}-1)}^{-1}]^{-1/2}

= [1-(x^{2}-1)/x^{2}]^{-1/2}

= [{x^{2}-(x^{2}-1)}/x^{2}]^{-1/2}

= [(x^{2}-x^{2}+1)/x^{2}]^{-1/2}

= [1/x^{2}]^{-1/2}

= [x^{2}]^{1/2}

= x

= [1-{(1-x

= [1-{-x

= [1-{x

= [1-(x

= [{x

= [(x

= [1/x

= [x

= x

3. a

[x^{l}/x^{m}]^{l2+lm+m2}.[x^{m}/x^{n}]^{m2+mn+n2}.[x^{n}/x^{l}]^{n2+nl+l2}

= [x^{(l-m)}]^{l2+lm+m2}.[x^{m-n}]^{m2+mn+n2}.[x^{n-l}]^{n2+nl+l2}

= x^{(l-m)(l2+lm+m2)}.x^{(m-n)(m2+mn+n2)}.x^{(n-l)(n2+nl+l2)}

Using a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2}), we get

= x^{l3-m3}.x^{m3-n3}.x^{n3-l3}

= x^{l3 - m3 + m3 - n3 + n3 - l3}

= x^{0} = 1

Hence, the simplified form of the given expression is 1

= [x

= x

Using a

= x

= x

= x

Hence, the simplified form of the given expression is 1

4. Using (a - b)

Given : x = p^{1/3} - p^{-1/3}

Take power3 on both the sides

x^{3} = (p^{1/3} - p^{-1/3})^{3}

Using (a-b)^{3} = a^{3} - b^{3} - 3ab(a-b)

x^{3} = (p^{1/3})^{3} - (p^{-1/3})^{3} - 3p^{1/3}.p^{-1/3}(p^{1/3}-p^{-1/3})

x^{3} = p - p^{-1} - 3p^{1/3 - 1/3}(x)

x^{3} = p - 1/p - 3p^{0}x

x^{3} = p - 1/p - 3(1)x

x^{3} = p - 1/p - 3x

x^{3} + 3x = p - 1/p

Take power3 on both the sides

x

Using (a-b)

x

x

x

x

x

x

5. On simplification,

is equal to

6. If x = 3

Given : x = 3^{1/3} + 3^{-1/3}

Take power3 on both the sides

x^{3} = (3^{1/3} + 3^{-1/3})^{3}

Using (a+b)^{3} = a^{3} + b^{3} + 3ab(a+b)

x^{3} = (3^{1/3})^{3} + (3^{-1/3})^{3} + 3.3^{1/3}.3^{-1/3}(3^{1/3}+3^{-1/3})

x^{3} = 3 + 3^{-1} + 3.3^{1/3 - 1/3}(x)

x^{3} = 3 + 1/3 + 3.3^{0}x

x^{3} = 3 + 1/3 + 3(1)x

x^{3} = 3 + 1/3 + 3x

x^{3} - 3x = 3 + 1/3

x^{3} - 3x = 10/3

3(x^{3} - 3x) = 10

3x^{3} - 9x = 10

Take power3 on both the sides

x

Using (a+b)

x

x

x

x

x

x

x

3(x

3x

7. If a

Given: a^{x} = b ----(1)

b^{y} = c ----(2)

c^{z} = a ----(3)

From the question a = c^{z}, b = a^{x}, c = b^{y}

Plug a = c^{z} in (1)

(c^{z})^{x} = b

c^{xz} = b

Now, plug c = b^{y}

(b^{y})^{xz} = b

b^{xyz} = b^{1}

Hence, the value of "xyz" is 1

b

c

From the question a = c

Plug a = c

(c

c

Now, plug c = b

(b

b

Hence, the value of "xyz" is 1

8. If 2

Let 2^{x} = 3^{y} = 6^{-z} = k

Then 2^{x} = k, 3^{y} = k, 6^{-z} = k

2 = k^{1/x}, 3 = k^{1/y}, 6 = k^{-1/z}

6 = k^{-1/z} ---> (2x3) = k^{-1/z} ---(1)

In (1), plug 2 = k^{1/x}, 3 = k^{1/y}

k^{1/x}.k^{1/y} = k^{-1/z}

k^{1/x + 1/y} = k^{-1/z}

1/x + 1/y = -1/z

1/x + 1/y + 1/z = 0

Then 2

2 = k

6 = k

In (1), plug 2 = k

k

k

1/x + 1/y = -1/z

1/x + 1/y + 1/z = 0

- Rationalization of surds
- Comparison of surds
- Operations with radicals
- Ascending and descending order of surds
- Simplifying radical expression
- Exponents and powers

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