Question

factorize (a+b)^3+64(a-b)^3

Answer 1

Answer:

Solution:-

x - 1 = 0

x = 1

Let p(x) = Ax³ + Bx² - 36x + 22

p(1) = A(1)³ + B(1)² - (36 × 1) + 22

⇒ A + B - 36 + 22 =0

⇒ A + B - 14 = 0

⇒ A + B = 14 ...........equation (1)

2^B = 64^A

⇒ 2^B = (2⁶)^A

⇒ B = 6A

⇒ A = B/6

Putting the value of A = B/6 in the equation, we get

B/6 + B = 14

⇒ B + 6B = 84

⇒ 7B = 84

⇒ B = 12

Putting the value of B = 12 in the equation A + B = 14, we get

A + 12 = 14

⇒ A =14 - 12

⇒ A = 2

Hence A = 2 and B = 12

Hence, x - 1 is a factor of 2x³ + 12x² - 36x + 22 and if we divide it by x - 1, we will get 2x² + 14x - 22 as quotient and 0 as remainder.

Let p(x) = 2x³ + 12x² - 36x + 22

p(1) = 2(1)³ + 12(1)² - (36 × 1) + 22

⇒ 2 + 12 - 36 + 22

⇒ 36 - 36 = 0

Answer 2

this is the answer

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