Solution
write the division as a fraction = 12y^4-4y^3 - 6y^2 + 36y - 22 over y-3 multiplied by r over y-3
r ×(2y^4 - 4y^3- 16y^2 + 36y- 22) over (y-3)^2
2ry^4 - 4ry^3 - 16ry^2 + 36ry - 22r over (y-3)^2
Why did you delete my answer
Answer:
Options (2) and (3)
Step-by-step explanation:
Let, 
-8 + 8i√3 = a² + b²i² + 2abi
-8 + 8i√3 = a² - b² + 2abi
By comparing both the sides of the equation,
a² - b² = -8 -------(1)
2ab = 8√3
ab = 4√3 ----------(2)
a = 
By substituting the value of a in equation (1),
48 - b⁴ = -8b²
b⁴ - 8b² - 48 = 0
b⁴ - 12b² + 4b² - 48 = 0
b²(b² - 12) + 4(b² - 12) = 0
(b² + 4)(b² - 12) = 0
b² + 4 = 0 ⇒ b = ±√-4
b = ± 2i
b² - 12 = 0 ⇒ b = ±2√3
Since, a =
For b = ±2i,
a = 
= 
= 
But a is real therefore, a ≠ ±2i√3.
For b = ±2√3
a = 
a = ±2
Therefore, (a + bi) = (2 + 2i√3) and (-2 - 2i√3)
Options (2) and (3) are the correct options.
Step-by-step explanation:
Left hand side:
4 [sin⁶ θ + cos⁶ θ]
Rearrange:
4 [(sin² θ)³ + (cos² θ)³]
Factor the sum of cubes:
4 [(sin² θ + cos² θ) (sin⁴ θ − sin² θ cos² θ + cos⁴ θ)]
Pythagorean identity:
4 [sin⁴ θ − sin² θ cos² θ + cos⁴ θ]
Complete the square:
4 [sin⁴ θ + 2 sin² θ cos² θ + cos⁴ θ − 3 sin² θ cos² θ]
4 [(sin² θ + cos² θ)² − 3 sin² θ cos² θ]
Pythagorean identity:
4 [1 − 3 sin² θ cos² θ]
Rearrange:
4 − 12 sin² θ cos² θ
4 − 3 (2 sin θ cos θ)²
Double angle formula:
4 − 3 (sin (2θ))²
4 − 3 sin² (2θ)
Finally, apply Pythagorean identity and simplify:
4 − 3 (1 − cos² (2θ))
4 − 3 + 3 cos² (2θ)
1 + 3 cos² (2θ)
