Rules of Indices .

Algebra Level 2

Simplify it and write the solution .

[ x b x c ] ( b + c − a ) × [ x c x a ] ( c + a − b ) × [ x a x b ] ( a + b − c ) \left[ \cfrac { x^{ b } }{ x^{ c } } \right] \overset { (b+c-a) }{ } \times \quad \left[ \frac { x^{ c } }{ x^{ a } } \right] \overset { (c+a-b) }{ } \times \quad \left[ \frac { x^{ a } }{ x^{ b } } \right] \overset { (a+b-c) }{ }


The answer is 1.

A Former Brilliant Member by A Former Brilliant Member

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2 solutions

Noel Lo
May 30, 2015

The power works out to be:

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

= b^2 - c^2 -a(b-c) + c^2 -a^2 - b(c-a) + a^2 - b^2 - c(a-b)

= -a(b-c) -b(c-a) - c(a-b) = -ab + ac -bc + ab - ca + bc = 0

x^0 = 1 unless x=0.

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E x p . = X b ( b + c − a ) + c ( c + a − b ) + a ( a + b − c ) X c ( b + c − a ) + a ( c + a − b ) + b ( a + b − c ) = X a 2 + b 2 + c 2 X a 2 + b 2 + c 2 = 1 Exp. = \dfrac{X^{b(b+c-a) + c(c+a-b) +a(a+b-c)} }{X^{c(b+c-a) +a(c+a-b) +b(a+b-c)} } \\ =\dfrac{X^{a^2+b^2+c^2}}{X^{a^2+b^2+c^2}} \\=1

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