If x^3-1=0 is a factor of x^6+ax^4+bx^3+cx^2+3x+2.

Then value of a^3+b^3+c^3=?

i think answer is   -36

if it is correct i will give explanation

i think answer is   -36

if it is correct i will give explanation

i dont know plz give the answer

 

By any chance are you asking for ?

x^3-1=0  is factor

thn  x-1   and  x^2+x+1  are factors

when we substitute 1 in equation we get  a+b+c=-6_____________1

and dividing given equation with x^2+x+1  we get remainder as

x(a-c+2) + (b-c+1)

but remainder shuld be zero   therefore  a-c+2=0_______2   and  b-c+1=0_________3

* x cannot be zero bcoz if x is zero and if we substitute 0 in equation we get

2=0  which is absurd

by solving 1,2,3 equations we get   a=-3   b=-2   c=-1therefore answer is -36

another method though lenghty,

since x3-1 is a factor ,by remainder theorem we get,

f(1)=0

f()=0

f(²)=0

add these three to get value of b.

we know that , a3+b3+c3-3abc=(a+b+c)(a+c2+b)(a2+c+b)

where the three expressions on the rhs u will get from the above three equations.

solve any two equations to get value of a,c,

put in above to get value.

i m getting ans as -54(by my method)

a=-3

b=-3

c=0

hey guys look at my method and tell me if i m wrong

x³-1 has 1 and w as its roots.

now putting these roots in the given eqn. we get

a+b+c=-6.....................................(1)

now putting w in the eqn we get

aw+b+1+cw²+3w+2=0

(3+b)+(a+3)w+cw²=0

also we know that 1+w+w²=0

therefore b+3=a+3=c............(2)

a=b=c-3

now from eqn i and 2

a+a+a=-9

a=-3,b=-3,c=0

ans=-54