Algebra Practice Questions Hard Level

Algebra questions basically involve modeling word problems into equations and then solving them. Some of the very basic formulae that come in handy while solving algebra questions are :

• (a + b) ² = a ² + b ² + 2 a b
• (a - b) ² = a ² + b ² - 2 a b
• (a + b) ² - (a - b) ² = 4 a b
• (a + b) ² + (a - b) ² = 2 (a ² + b ² )
• (a² - b² ) = (a + b) (a - b)
• (a + b + c) ² = a ² + b ² + c ² + 2 (a b + b c + c a)
• (a ³ + b ³ ) = (a + b) (a ² - a b + b ² )
• (a ³ - b ³ ) = (a - b) (a ² + a b + b ² )
• (a³ + b³ + c³ - 3 a b c) = (a + b + c) (a² + b² + c² - a b - b c - c a)
• If a + b + c = 0, then a³ + b³ + c³ = 3 a b c
• For a quadratic equation ax² + bx + c = 0, x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Check: Tips & Tricks for Algebra

Solved Questions on Algebra (Hard)

Question 1: 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: 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 

Practice Problems on Algebra (Hard)

Question 1: If a = 1 − 1/b and b = 1 − 1/c​, find the value of c − 1/a.

Question 2: If a + b + c = 3, find the value of: 1/(1 − a)(1 − b)​ + 1/(1 − b) (1 − c) ​+ 1/(1 − c) (1 − a).

Question 3: If a + 1/a = √3, calculate a¹⁸ + a¹² + a⁶ + 1.

Question 4: If x = 8, calculate x⁵ − 9x⁴ + 9x³ − 9x² + 9x − 1.

Question 5: If x² + y² + z² = 2(x + y − 1), calculate x³ + y³ + z³.

Question 6: If x¹² + 1 /x⁶ = 6, find the value of x³⁶ + 1/x¹⁸.

Question 7: If x + y + z = 6, x² + y² + z² = 26, and xy + yz + zx = 14, find the value of x³ + y³ + z³ − 3xyz.

Question 8: if a = 2 + √3 and b = 2 − √3, find the value of: a⁵ - b⁵/a - b.

Answer Key:

1. b + 1​
2. 0
3. 0
4. 7
5. 2
6. 198
7. 36
8. 82

Practice More –

• Algebra Practice Questions Easy Level
• Algebra Practice Questions Medium Level

Quiz – Algebra Quiz