# Algebra | Set-2

**Question 1:** If x + 1/x = -2, then the value of x^{1000} + 1/x^{1000} is**Solution.** x= -1 satisfy the above eq.

So, (-1)^{1000} + 1/(-1)^{1000} = 1 + 1 = **2**

**Question 2:** If a = 1 – 1/b and b = 1 – 1/c, then the value of c – 1/a is**Solution.** Let a = 2 and b = -1

2 = 1 – (-1)

2 = 2 True

So, find the value of c

b = 1 – 1/c

-1 = 1 – 1/c

c =1/2

Then, c – 1/a = 1/2 – 1/2 = **0**

**Question 3:** If a + b + c = 3, then the value of 1/(1 – a)(1 – b) + 1/(1 – b)(1 – c) + 1/(1 – c)(1 – a)**Solution** 1/(1 – a)(1 – b) + 1/(1 – b)(1 – c) + 1/(1 – c)(1 – a)

=> [(1 – c) + (1 – a) + (1 – b)]/(1 – a)(1 – b)(1 – c)

=> [3 – (a + b + c)]/(1 – a)(1 – b)(1 – c)

=> 3 – 3 /(1 – a)(1 – b)(1 – c)

=> **0**

**Question 4:** If a + 1/a = √3, then the value of a^{18} + a^{12} + a^{6} + 1 is**Solution.** a^{3} + 1/a^{3} = (a + 1/a)^{3} – 3(a + 1/a)

=> 3 √3 – 3 √3

=> 0

a^{3} + 1/a^{3} = 0

a^{6} + 1 = 0

Then,

a^{18} + a^{12} + a^{6} + 1

a^{12}(a^{6} + 1) + (a^{6} + 1)

a^{12} x 0 + 0 = **0**

**Question 5:** If a = √3 + 1 / √3 -1 and b = √3 -1 / √3 + 1, then find the value of

(a^{2} + ab + b^{2})/(a^{2} – ab + b^{2}) is**Solution.** a = 1/b

therefore ab = 1

a + b = (√3 + 1) / ( √3 -1) + (√3 -1) / (√3 + 1)

=> (3 + 1 + 2√3 + 3 + 1 – 2√3)/ (3 – 1)

=> 8/2

=> 4

a + b = 4

a^{2} + b^{2} = 4^{2} – 2 *(ab)

a^{2} + b^{2} = 14

Now, (a^{2} + ab + b^{2})/(a^{2} – ab + b^{2})

=>(14 + 1)/(14 -1)

=> **15/13**

**Question 6:** If x = 8, then find value of x^{5} – 9x^{4} + 9x^{3} – 9x^{2} + 9x^{1} – 1**Solution.** We can write it as

8^{5} – 8*x^{4} – 1*x^{4} + 8*x^{3} + 1*x^{3} – 8*x^{2} – 1*x^{2} +8*x^{1}+ 1*x^{1} – 1

Now put x = 8

8^{5} – 8*8^{4} – 1*8^{4} + 8*8^{3} + 1*8^{3} – 8*8^{2} – 1*8^{2} +8*8^{1}+ 1*8^{1} – 1

= 8 – 1

= **7**

**Question 7:** If the sum of the square of two real numbers is 74 and their sum is 12. Then the sum of cubes of these two numbers is**Solution** Let two numbers are a and b

Given, a^{2} + b^{2} = 74

a + b = 12

(a + b)^{2} = a^{2} + b^{2} + 2ab

12^{2} = 74 + 2ab

144 = 74 + 2ab

ab = 35

We get a=7 and b=5

Then, a^{3} + b^{3} = 7^{3} + 5^{3}

=343 + 125

=**468**

**Question 8:** If ab(a+b)=m, then the value of a^{3} + b^{3} + 3 m is**Solution.** ab(a + b)=m

a + b = m/ab

Cubing both sides

a^{3} + b^{3} + 3ab(a + b) = m^{3} / a^{3}b^{3}

a^{3} + b^{3} + 3 m = **m ^{3} / a^{3}b^{3}**

**Question 9:** If m=√7+√7+√7….. and n=√7-√7-√7…….

then among the following relation between m and n holds is**Solution.** m = √(7 + m)

m^{2} = 7 + m

m^{2} – m = 7…….(1)

and n = √(7 – n)

n^{2} + n = 7…….(2)

from (1) and (2)

m^{2} – m = n^{2} + n

m^{2} – n^{2} – (m + n) = 0

(m + n)(m – n) – (m + n)= 0**m – n – 1 = 0**

**Question 10:** If x^{2} + y^{2} + z^{2} = 2(x + y -1), then the value of x^{3} + y^{3} + z^{3}?**Solution.** x^{2} + y^{2} + z^{2} = 2x + 2y -2

(x^{2} + 1 -2x) +(y^{2} + 1 -2y) + (z^{2}) = 0

(x – 1)^{2} + (y – 1)^{2} + (z)^{2} = 0

=> (x – 1)^{2} = 0

=> x = 1

(y – 1)^{2} = 0

=> y=1

(z)^{2} = 0

=> z = 0

Put value in eq

x^{3} + y^{3} + z^{3}

1^{3} + 1^{3} + 0^{3}

=> **2**

**Question 11:** If (x^{12} + 1 )/x^{6} = 6, then the value of (x^{36} + 1 )/x^{18} ?**Solution.** Given

(x^{12} + 1 )/x^{6} = 6

x^{6} + 1 /x^{6} = 6

Cubing both sides

(x^{6} + 1 /x^{6})^{3} = 6^{3}

x^{18} + 1/x^{18} + 3 (x^{6} + 1 /x^{6}) = 216

x^{18} + 1/x^{18} + 3 * 6 = 216

x^{18} + 1/x^{18} = 198

(x^{36} + 1)/x^{18} = **198**