Question

a,b,c are positive real numbers such that a² + b² = c² and ab = c. Determine the value of

|(a+b+c)(a+b-c)(b+c-a)(c+a-b)|
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c²​

Answer 1

Step-by-step explanation:

We know 2(ab+ac+bc)=(a+b+c)

2

−a

2

−b

2

−c

2

2(ab+ac+bc)=(a+b+c)

2

−1

(ab+bc+ac)=

2

(a+b+c)

2

−1

Now if a=b=c=

3

1

, then the above expression attains maximum value of 1.

Hence for distinct values of a,b and c,

2

(a+b+c)

2

−1

<1

Hence, ab+bc+ac<1

Answer 2

Correct Question:

a,b,c are positive real numbers such that a²+b²=c² and ab=c. Determine the value of |(a+b+c)(a+b-c)(b+c-a)(c+a-b)/c²|

Step by step Answer:

Given:

a²+b²=c²

ab=c

To find:

|(a+b+c)(a+b-c)(b+c-a)(c+a-b)/c²|

First part:

(a+b+c)(a+b-c)

=[(a+b)+c][(a+b)-c]

=(a+b)²-c²

=a²+2ab+b²-c²

Second part:

(b+c-a)(c+a-b)

=bc + ab - b² + c²+ ac - bc - ac - a² + ab

=2ab-b²+c²-a²

Third part:

(a²+2ab+b²-c²)(2ab-b²+c²-a²)/c²

=(a²+b²+2ab-c²)(-a²-b²+2ab+c²)/c²

=(c²+2c-c²)(-c²+2c+c²)/c² [as a²+b²=c² and ab=c]

=2c×2c/c²

=4c²/c²

=4

Final answer:

| 4 | = 4