Test: SSC CGL Previous Year Questions: Algebra (2023-18) - 3 - SSC CGL MCQ

x³ + y³ = (x + y) (x² + y² - xy)
= (x + y) [(x + y)² - 3xy]
405 = 9(81- 3ab) ⇒ 45 = 81 - 3ab
3ab = 36 ⇒ ab = 12

(2x - (3/x)) = 2
Squaring both side
4x² + 9/(x²) - 2 × 2 × 3 = 4
4x² + 9/(x²) = 16 (squaring both side)
(16x⁴ + 81/(x⁴)) + 2 × 4 × 9 = 256
(16x⁴ + 81/(x⁴)) = 256 - 72 = 184

a - (12/a) = 1
a² -12 = a ⇒ a² - a - 12 = 0
a² - 4a + 3a -12 = 0
a(a - 4) + 3(a - 4) = 0
(a + 3)(a - 4) = 0 ⇒ a = - 3 and 4
Put a = 4
a² + 16/a² = 16 + 16/16 = 16 + 1 = 17

4x⁴ - 37x² + 9 = 0 dividing both side by x²
Then 4x² + 9/x² = 37 subtracting 12 and taking root from both side
2x - 3/x = 5 taking cube on both side
8x³ - 27/(x³) = 5³ + 3 × 5 × 2 × 3 = 125 + 90 = 215

a² + c² + 17 = 2(a - 8b - 2b²)
a² + c² + 17 - 2a + 16b + 4b² = 0
(a - 1)² + (2b + 4)² + c² = 0
a = 1, b = - 2, c = 0
a³ + b³ + c³ = (1)³+ (-2)³ + 0 = - 7

Put a = 2, b = -1,c = -1

a² + b² + c² + 216 = 12(a + b - 2c)
(a − 6)² + (b - 6)² + (c + 12)² = 0
a = 6, b = 6, c = - 12

= √180 = 6√5

(56√7x³ − 2√2y³) ÷ (2√7x − √2y)
= Ax² + By² − Cxy,

= Ax² + By² − Cxy,

Comparing the above equation we get,
A = 28, B = 2, C = -2√14
A + B − √14C = 28 + 2 + 28 = 58

x = 7, y = 3 satisfies given equation
x⁴ - y⁴ = 2401 - 81 = 2320

8 + 2px² - 36x - 27x³ = (2 - 3x)³
8 + 2px² - 36x - 27x³ = 8 + 54x² - 36x - 27x³
Comparing coefficient of x² term, p = 27

Squaring both side,

Short-Trick: 

x = 4, satisfy

y = 2x + 1,
y - 2x = 1
Cubing both sides,
y³ - 8x³ - 3 × y × 2x (y - 2x) = 1
y³ - 8x³ - 6xy = 1
8x³ - y³ + 6x y = -1

Taking the square of (3x + y)²
(3x + y)² = 9x² + y² + 6xy
(3x + y)² = 37 + 6 × 2
(3x + y)² = 49 ⇒ 3x + y = 7
Taking cube on both side
27x³ + y³ + 9xy(3x + y) = 343
27x³ + y³ = 343 - 9 x 2 x 7
27x³ + y³ = 217

÷ 

= 0.75 + 0.25 = 1

on rationalization,

= ( 20 - 16 ) ÷ 4 = 1

x + y = 8, y + z = 13 and z + x = 17
On adding above equations,
2(x + y + z) = 38 ⇒ x + y + z = 19
x = 6, y = 2, z = 11

(a + b - c)³ + (a - b + c)³ - 8a³
⇒ (a + b - c)³ + (a - b + c)³ + ( - 2a)³
We know that if x + y + z = 0
Then x³ + y³ + z³ = 3xyz
Here, x = a + b - c ,y = a - b + c , z = - 2a
So, (a + b - c)³ + (a - b + c)³ - 8a³
= 6a(a - b + c)(c - a - b)
Short trick :
Put a = 1 and b = c = 0
(a + b - c)³ + ( a - b + c)³ - 8a³
⇒ (1 + 0 - 0)³ + (1 - 0 + 0)³ - 8(1)³
= 1 + 1 - 8 = - 6
Now on checking the options, option (a) is satisfied.
6a(a - b + c)(c - a - b)
⇒ 6(1)(1 - 0 + 0)(0 - 1 - 0) = - 6
condition satisfied hence option (a) is the right answer.

In the equation:
(a + b - c + d)² - (a - b + c - d)²
Use formula : x² - y² = (x - y)(x + y)
(a + b - c + d)² - (a - b + c - d)² = (a + b - c + d + a - b + c - d)(a + b - c + d - a + b - c + d)
= (2a)[2(b - c + d)]
= 4a(b - c + d)

(P ÷ Q) x R = 

= 2 (x² + y²)

a³ + 1/(a³) = 52
⇒ a + 1/a = 4 ⇒ 2(a + 1/a) = 8

if x + y + z = 0, then x³ + y³ + z³ = 3xyz
Similarly, a² + b² - c² = 0
Therefore, a⁶ + b⁶ - c⁶ = 3a² b² c²

a = 2b = 8c
a : b : c = 8 : 4 : 1
a + b + c = 13
Thus, a = 8; b = 4 and c = 1
 = 9/2

x + y = 13, y + z = 15, z + x = 16
⇒ 2(x + y + z) = 13 + 15 + 16 = 44
x + y + z = 22
x = 7, y = 6, z = 9
15/54 = 5/18

Given,

Multiply and divide by 2


Now, 

Put a = 2 , b = 1 and c = 0
ab(a - b) + bc(b - c) + ca(c - a) = 2 × 1(2 - 1) + 1 × 0 (1 - 0) + 0 × 2 (0 - 2) = 2
Now, Go throw options Choose option (d)
⇒ (b - a)(b - c)(c - a)
= (1 - 2)(1 - 0)(0 - 2) = 2
Condition satisfied.
Similarly you can put these values in other options but only option (d) will satisfy the condition.

Given, 

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
⇒ x² + y² + z² = 361 − 228 = 133

Given,
x⁴ - 6x² - 1 = 0, x² - 1/(x²) = 6
Cubing both sides

 ...[(a - b)³ = a³ - b³ - 3ab(a - b)]
⇒ x⁶ - 1/(x⁶) = 216 + 18 = 234

⇒ 234 - 5(6) + 5 = 209