As we know from the formula,
A³ + B³ + C³ = ( A + B + C) (A² + B² + C² – AB – BC – CA) + 3ABC
If A + B + C = 0 then,
A³ + B³ + C³ = 0 x (A² + B² + B² – AB – BC – CA) + 3ABC
A³ + B³ + C³ = 0 + 3ABC
A³ + B³ + C³ - 3ABC= 0

Correct Option: B

As we know from the formula,
A³ + B³ + C³ = ( A + B + C) (A² + B² + C² – AB – BC – CA) + 3ABC
If A + B + C = 0 then,
A³ + B³ + C³ = 0 x (A² + B² + B² – AB – BC – CA) + 3ABC
A³ + B³ + C³ = 0 + 3ABC
A³ + B³ + C³ - 3ABC= 0
As we know that from given question,
A = a, B = b and C = 1, Put these value in above equation, we will get
a³ + b³ + 1³ - 3 x a x b x 1 = 0
a³ + b³ + 1 - 3ab = 0

When p is wrong i.e., -b/a = (α + β) is wrong but c/a =(αβ) is correct.
Then αβ = c/a = 2 x 6 = 12 ....(i)

Again, when q is wrong i.e.,c/a = (α.β ) is wrong but -b/a = α+β is correct.
Then, -b/a = α+β= 2 + (-9) = -7

Correct Option: C

When p is wrong i.e., -b/a = (α + β) is wrong but c/a =(αβ) is correct.
Then αβ = c/a = 2 x 6 = 12 ....(i)

Again, when q is wrong i.e.,c/a = (α.β ) is wrong but -b/a = α+β is correct.
Then, -b/a = α+β= 2 + (-9) = -7

So, the required correct quadratic equations is
x² -(α + β)x + α.β=0
⇒ x² - (-7)x + 12 = 0
⇒ x² + 7x + 12 = 0
and correct roots this equations are -3, -4.

Let the cost of each apple,orange nd pear be Rs.x,y and z, respectively.Then,
2x + 3y + z = 62 ...(i)
5x + 6y + 4z = 20 ...(ii)

Correct Option: C

Let the cost of each apple,orange nd pear be Rs.x,y and z, respectively.Then,
2x + 3y + z = 62 ...(i)
5x + 6y + 4z = 20 ...(ii)
On subtracting Eq.(i)from Eq.(ii), we get
3x + 3y + 3z = 58
So, the cost of 3 apple,3 oranges and 3 pears is Rs. 58.