If #p(x)=x^3-3x^2+2x+5 and p(a)=p(b)=p(c)# then the numerical value of #(2-a)(2-b)(2-c)=#?

2 Answers
George C.
Jul 17, 2018

If #p(a)=p(b)=p(c)=k# then:

#(2-a)(2-b)(2-c) = 5-k#

Explanation:

Given:

#p(x) = x^3-3x^2+2x+5#

Suppose:

#p(a) = p(b) = p(c) = k#

where #a, b, c# are all distinct.

Then:

#0 = x^3-3x^2+2x+(5-k)#

#color(white)(0) = (x-a)(x-b)(x-c)#

#color(white)(0) = x^3-(a+b+c)x^2+(ab+bc+ca)x-abc#

So:

#{ (a+b+c=3), (ab+bc+ca=2), (abc = k-5) :}#

We find:

#(2-a)(2-b)(2-c) = 8-4(a+b+c)+2(ab+bc+ca)-abc#

#color(white)((2-a)(2-b)(2-c)) = 8-4(3)+2(2)-(k-5)#

#color(white)((2-a)(2-b)(2-c)) = 8-12+4-k+5#

#color(white)((2-a)(2-b)(2-c)) = 5-k#

So if the question should have specified #p(a)=p(b)=p(c)color(red)(=0)#, then the answer would be #(2-a)(2-b)(2-c) = 5#

maganbhai P.
Jul 17, 2018

# (2-a)(2-b)(2-c)=-abc#
There seems to be something missing from the question,

Explanation:

Here, #p(x)=x^3-3x^2+2x+5 and p(a)=p(b)=p(c)#

#:.p(a)=p(b)=>a^3-3a^2+2a+5=b^3-3b^2+2b+5#

#=>(a^3-b^3)-3(a^2-b^2)+2(a-b)=0#

#=>a^2+ab+b^2-3a-3b+2=0to#[dividing by #a-b!=0]#

#p(a)=p(b)=>a^2+b^2+ab-3a-3b+2=0to(1)#
Similarly,

#p(b)=p(c)=>b^2+c^2+bc-3b-3c+2=0to(2)#

#p(c)=p(a)=>c^2+a^2+ca-3c-3a+2=0to(3)#

#(2)-(3)=>b^2-a^2+c(b-a)-3(b-a)=0#

#=>color(red)(a+b+c=3to(4)#

#=>(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca=(3)^2#

#=>a^2+b^2+c^2+2ab+2bc+2ca=9to(5)#

Now , taking #(1)+(2)+(3)#

#2a^2+2b^2+2c^2+ab+bc+ca-6(a+b+c)+6=0#

#=>2a^2+2b^2+2c^2+ab+bc+ca-6(3)+6=0to[use(4)]#

#=>2a^2+2b^2+2c^2+ab+bc+ca=12to(6)#

Taking , eqn.#(6)-2(5)# ,we get

#color(white)(....)2a^2+2b^2+2c^2+ab+bc+ca=12#

#ul(-2a^2-2b^2-2c^2-4ab-4bc-4ca=-18)#
#=>-3ab-3bc-3ca=-6=>color(red)(ab+bc+ca=2to(7)#

From #(5)and(7)# we get

#a^2+b^2+c^2+2(2)=9=>color(red)(a^2+b^2+c^2=5to(8)#

Now,

#a^3+b^3+c^3#=#(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc#

#a^3+b^3+c^3=3(5-2)+3abcto[use ,(4),(7)and(8)]#

#=>a^3+b^3+c^3=9+3abcto(9)#

If we take ,

#(a+b+c)^3#=#a^3+b^3+c^3+3[(a+b+c)(ab+bc+ca)-abc]#

#color(red)("Which gives the same result as eqn.(9) "#
#color(red)("and we can not find any one of "# #color(red)(a^3+b^3+c^3 or abc#

Let us take ,

#X=(2-a)(2-b)(2-c)#

#=>X=(2-a)(4-2b-2c+bc)#

#=>X=8-4b-4c+2bc-4a+2ab+2ca-abc#

#=>X=8-4(a+b+c)+2(ab+bc+ca)-abc#

#=>X=8-4(3)+2(2)-abc#

#=>X=8-12+4-abc#

#=>X=color(red)(-abcto "Undetermined"#

#i.e. (2-a)(2-b)(2-c)=color(red)(-abc)#

I agree with @George C. "There seems to be something missing from the question ."

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