Math Olympiad Test: Number System- 1 - Class 9 MCQ

If x + y = 12 and xy = 35 then
(x + y)² = (12)²
Now x² + y² = (x + y)² - 2xy
⇒ x² + y² = 144 - 2(xy) = 144 - 2(35)
= 144 - 70 = 74

Here 991 × 1009 = (1000 - 9) × (1000 + 9) = (1000)² - (9)
= (10)⁶ - 81 = 999919

(x - y + z)² will contain 6 terms in its simplied form.

Here (α + b + c)² = (α² + b² + c²)
+ 2 (αb + bc + cα)
⇒ (a + b + c)² = (16) + 2(10) = 16 + 20 = 36
⇒ (α + b + c) = 36 = ±6

Expanding the left side of the equation using the binomial cube formula, we have:

(x + k)³ + (x - k)³ = 2x³ + 54x

(x³ + 3x²k + 3xk² + k³) + (x³ - 3x²k + 3xk² - k³) = 2x³ + 54x

2x³ + 6xk² = 2x³ + 54x

Canceling out the common terms on both sides, we get:

6xk² = 54x

Dividing both sides of the equation by 6x, we have:

k² = 9

Taking the square root of both sides, we get:

k = ±3

Therefore, k can be either 3 or -3.

Hence, the correct answer is option 1: k = 3, -3.

Let α = 30, b = 20, c = -50
∵ α + b + 3 = 0
∴ α³ + b³ + c³ = 3αbc
= 3(30)(20)(-50)
= - 90,000

We have 


∴ Cube root will be 

Here 3x² - 27 = 3(x² - 9)
= 3[(x)² - (3)²]
= 3(x + 3) (x - 3)

We have α³ + b³ + c³ - 3αbc
= (α + b + c)(α² + b² + c² - αb - bc - cα)
If  (α + b + c) ≠ 0 then α³ + b³ + c³ - 3αbc will posses 0 value, if α² + b² + c² - αb - bc - cα = 0, and this will only happen,
when α = b = c

[Since, (a)³ + (b)³ = (a + b) (a² - ab + b²)]

⇒ (0.013 + 0.007)

⇒ 0.02