Question

(bc)^b-c × (ca)^c-a × (ab)^a-b/ (a^b-c×b^c-a× c^a-b)^-1​

Answer 1

Before we solve

The question requires one of the exponent laws. If a number is in the denominator, we add the negative sign to the exponent.

The hint is in the denominator. Let's see how we can simplify this fraction.

Solution

Given

$\dfrac{(bc)^{b-c}\cdot (ca)^{c-a}\cdot(ab)^{a-b}}{(a^{b-c}\cdot b^{c-a}\cdot c^{a-b})^{-1}}$

$=(bc)^{b-c}\cdot (ca)^{c-a}\cdot(ab)^{a-b}\div (a^{b-c}\cdot b^{c-a}\cdot c^{a-b})^{-1}$

$=(bc)^{b-c}\cdot (ca)^{c-a}\cdot(ab)^{a-b}\cdot a^{b-c}\cdot b^{c-a}\cdot c^{a-b}$

$=a^{c-a+a-b+b-c}\cdot b^{b-c+a-b+c-a}\cdot c^{b-c+c-a+a-b}$

$=a^{0}\times b^{0}\times c^{0}$

$=\boxed{1}$

So, the answer is $\boxed{1}$.

More Information

Intuitions behind Exponent Laws

• Fractional powers

→ Applying the same power on both sides keeps both equal. Since applying natural power to the number removes the power, we say $\sqrt[n]{x} =x^{\frac{1}{n} }$.

• Negative powers

→ Applying negative power for denominator comes from fraction reduction. Since applying it removes the power, we say $x^{-n}=\dfrac{1}{x^{n}}$.

Answer 2

Answer

1

Step-by-Step Explanation

We expand the terms in the numerator by using an exponent rule

$({b}^{b - c} . {b}^{a - b} . {c}^{b - c} . {c}^{c - a} . {a}^{c - a} . {a}^{a - b}) \div ( {a}^{b - c} . {b}^{c - a} . {c}^{a - b})^{ - 1}$

We then get

[a^(c-b) * b^(a-c) * c^(b-a)] * [a^(b-c) * b^(c-a) * c^(a-b)]

The terms with the same bases are the reciprocal of each other and hence they cancel out

Thus we get 1.