if the zeroes of the polynomial, p(x)= ax^3 + 3bx^2 + 3cx + d = 0 be alpha - beta, alpha and alpha + beta. Prove that 3abc = a^2d + 2b^3 ...I hope you help..

Step-by-step explanation:

According to question

α−d+α+α+d=−3ba

3α=−3ba

α=−ba

α2−dd+α2+αd+α2−d2=3a

3α2−d2=3ca

3b2a2−h2=3ca

d2=3b2a2−3ca=3b2−3aca2

α(α2−d2)=−da

−ba(b2a2−(3b2−3aca2))=−da

b(b2−3b2+3aca2)=d

−2b3+3ab=a2d

2b3−3abc+a2d=0

pls mark me as brainliest..

the answer is the option b

the range is all real numbers

step-by-step explanation:

we know that

the absolute value parent function is equal to

see the attached figure to better understand the problem

so

a. it is v-shaped > is true

b. the range is all real numbers > is not true

because the range is the > [0,∞)

all real numbers greater than or equal to zero

c. it goes through the origin > is true

d. the domain is all real numbers > is true

step-by-step explanation:

the anwser is d) 75°

step-by-step explanation: