Algebra | Set-2

Question 1: If x + 1/x = -2, then the value of x¹⁰⁰⁰ + 1/x¹⁰⁰⁰ 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.

a = 1 – 1/b

=>ab=b-1

=>1/a=b/(b-1) ——–(1)

And

 b =1-1/c =>

b + 1/c = 1

=> bc + 1 = c

=> bc – c  =  -1

=> c(b – 1)  = -1

=> c  = 1/(1 – b) ———–(2)

putting the values of 1/a and c from above 1 and 2 in c – 1/a,

1/(1 – b)- b/(b-1) =    (b+1)/(1-b)                                                            

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 a¹⁸ + a¹² + a⁶ + 1 is 
Solution. a³ + 1/a³ = (a + 1/a)³ – 3(a + 1/a) 
=> 3 √3 – 3 √3 
=> 0 
a³ + 1/a³ = 0 
a⁶ + 1 = 0 
Then, 
a¹⁸ + a¹² + a⁶ + 1 
a¹²(a⁶ + 1) + (a⁶ + 1) 
a¹² x 0 + 0 = 0

Question 5: If a = √3 + 1 / √3 -1 and b = √3 -1 / √3 + 1, then find the value of 
(a² + ab + b²)/(a² – ab + b²) 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 
a² + b² = 4² – 2 *(ab) 
a² + b² = 14 
Now, (a² + ab + b²)/(a² – ab + b²) 
=>(14 + 1)/(14 -1) 
=> 15/13

Question 6: If x = 8, then find value of x⁵ – 9x⁴ + 9x³ – 9x² + 9x¹ – 1 
Solution. We can write it as 
8⁵ – 8*x⁴ – 1*x⁴ + 8*x³ + 1*x³ – 8*x² – 1*x² +8*x¹+ 1*x¹ – 1 
Now put x = 8 
8⁵ – 8*8⁴ – 1*8⁴ + 8*8³ + 1*8³ – 8*8² – 1*8² +8*8¹+ 1*8¹ – 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, a² + b² = 74 
a + b = 12 
(a + b)² = a² + b² + 2ab 
12² = 74 + 2ab 
144 = 74 + 2ab 
ab = 35 
We get a=7 and b=5 
Then, a³ + b³ = 7³ + 5³
=343 + 125 
=468

Question 8: If ab(a+b)=m, then the value of a³ + b³ + 3 m is 
Solution. ab(a + b)=m 
a + b = m/ab 
Cubing both sides 
a³ + b³ + 3ab(a + b) = m³ / a³b³
a³ + b³ + 3 m = m³ / a³b³

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) 
m² = 7 + m 
m² – m = 7…….(1) 
and n = √(7 – n) 
n² + n = 7…….(2) 
from (1) and (2) 
m² – m = n² + n 
m² – n² – (m + n) = 0 
(m + n)(m – n) – (m + n)= 0 
m – n – 1 = 0

Question 10: If x² + y² + z² = 2(x + y -1), then the value of x³ + y³ + z³? 
Solution. x² + y² + z² = 2x + 2y -2 
(x² + 1 -2x) +(y² + 1 -2y) + (z²) = 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 
x³ + y³ + z³
1³ + 1³ + 0³
=> 2

Question 11: If (x¹² + 1 )/x⁶ = 6, then the value of (x³⁶ + 1 )/x¹⁸ ? 
Solution. Given 
(x¹² + 1 )/x⁶ = 6 
x⁶ + 1 /x⁶ = 6 
Cubing both sides 
(x⁶ + 1 /x⁶)³ = 6³
x¹⁸ + 1/x¹⁸ + 3 (x⁶ + 1 /x⁶) = 216 
x¹⁸ + 1/x¹⁸ + 3 * 6 = 216 
x¹⁸ + 1/x¹⁸ = 198 
(x³⁶ + 1)/x¹⁸ = 198