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abc = 8, a^2 +b^2 +c^2 =21, a+b+c=1, find (1-a)(2-b)(3-c)

abc = 8, a^2 +b^2 +c^2 =21, a+b+c=1, find (1-a)(2-b)(3-c)

Grade:11

2 Answers

Deepak Kumar Shringi
askIITians Faculty 4405 Points
3 years ago
a=4 ,b=-2 , c=-1
put values to get
(1-a)(2-b)(3-c)
(-3)(4)(4)
= -48
Aditya Gupta
2080 Points
3 years ago
the correct method to approach this ques is as follows:
a+b+c=1
squaring
21+2(ab+bc+ca)=1
or ab+bc+ca= – 10
now consider the equation with roots a, b, c
x^3 – (a+b+c)x^2+(ab+bc+ca)x – abc=0
or x^3 – x^2 – 10x – 8=0
or (x-4)*(x+1)*(x+2)=0
so {a, b, c} belong to {4, -2, -1}.
now you can proceed further. note that (1-a)(2-b)(3-c) will not have a unique value but depend upon what values a b and c are assigned.

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