Solveeit Logo
Login

Question

Question: The numerical value of $(x^{1/a-b})^{1/a-c} \times (x^{1/b-c})^{1/b-a}$ $\times (x^{1/c-a})^{1/c-b}$...

The numerical value of (x1/a−b)1/a−c×(x1/b−c)1/b−a(x^{1/a-b})^{1/a-c} \times (x^{1/b-c})^{1/b-a} ×(x1/c−a)1/c−b\times (x^{1/c-a})^{1/c-b} is (a,b,c are distinct real numbers)(a, b, c \text{ are distinct real numbers})

A

1

B

8

C

0

D

None

Answer

1

Explanation

Solution

The problem asks for the numerical value of the given expression: (x1/(a−b))1/(a−c)×(x1/(b−c))1/(b−a)×(x1/(c−a))1/(c−b)(x^{1/(a-b)})^{1/(a-c)} \times (x^{1/(b-c)})^{1/(b-a)} \times (x^{1/(c-a)})^{1/(c-b)} where a,b,ca, b, c are distinct real numbers.

We will use the following exponent rules:

  1. (pm)n=pmn(p^m)^n = p^{mn}
  2. pm×pn=pm+np^m \times p^n = p^{m+n}
  3. p0=1p^0 = 1 (for p≠0p \neq 0)

Step 1: Simplify each term using the rule (pm)n=pmn(p^m)^n = p^{mn}.

First term: (x1/(a−b))1/(a−c)=x1(a−b)(a−c)(x^{1/(a-b)})^{1/(a-c)} = x^{\frac{1}{(a-b)(a-c)}}

Second term: (x1/(b−c))1/(b−a)=x1(b−c)(b−a)(x^{1/(b-c)})^{1/(b-a)} = x^{\frac{1}{(b-c)(b-a)}}

Third term: (x1/(c−a))1/(c−b)=x1(c−a)(c−b)(x^{1/(c-a)})^{1/(c-b)} = x^{\frac{1}{(c-a)(c-b)}}

Step 2: Combine the terms using the rule pm×pn=pm+np^m \times p^n = p^{m+n}.

The entire expression becomes: x1(a−b)(a−c)+1(b−c)(b−a)+1(c−a)(c−b)x^{\frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-b)}}

Step 3: Simplify the sum of the exponents.

Let EE be the exponent: E=1(a−b)(a−c)+1(b−c)(b−a)+1(c−a)(c−b)E = \frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-b)}

To simplify this sum, we can rewrite the terms in the denominators to have a consistent order, for example, using (a−b)(a-b), (b−c)(b-c), and (c−a)(c-a).

Notice the following relationships: b−a=−(a−b)b-a = -(a-b) a−c=−(c−a)a-c = -(c-a) c−b=−(b−c)c-b = -(b-c)

Now, substitute these into the terms of EE: The first term remains: 1(a−b)(a−c)\frac{1}{(a-b)(a-c)} The second term becomes: 1(b−c)(−(a−b))=−1(b−c)(a−b)\frac{1}{(b-c)(-(a-b))} = \frac{-1}{(b-c)(a-b)} The third term becomes: 1(c−a)(−(b−c))=−1(c−a)(b−c)\frac{1}{(c-a)(-(b-c))} = \frac{-1}{(c-a)(b-c)}

So, E=1(a−b)(a−c)−1(b−c)(a−b)−1(c−a)(b−c)E = \frac{1}{(a-b)(a-c)} - \frac{1}{(b-c)(a-b)} - \frac{1}{(c-a)(b-c)}

To add these fractions, we find the least common multiple (LCM) of the denominators. The LCM of (a−b)(a−c)(a-b)(a-c), (b−c)(a−b)(b-c)(a-b), and (c−a)(b−c)(c-a)(b-c) is (a−b)(b−c)(c−a)(a-b)(b-c)(c-a).

Let's rewrite the first term using (c−a)(c-a) instead of (a−c)(a-c): 1(a−b)(−(c−a))=−1(a−b)(c−a)\frac{1}{(a-b)(-(c-a))} = \frac{-1}{(a-b)(c-a)}.

Now, express all terms with the common denominator (a−b)(b−c)(c−a)(a-b)(b-c)(c-a): E=−1(a−b)(c−a)×(b−c)(b−c)−1(b−c)(a−b)×(c−a)(c−a)−1(c−a)(b−c)×(a−b)(a−b)E = \frac{-1}{(a-b)(c-a)} \times \frac{(b-c)}{(b-c)} - \frac{1}{(b-c)(a-b)} \times \frac{(c-a)}{(c-a)} - \frac{1}{(c-a)(b-c)} \times \frac{(a-b)}{(a-b)}

E=−(b−c)(a−b)(b−c)(c−a)−(c−a)(a−b)(b−c)(c−a)−(a−b)(a−b)(b−c)(c−a)E = \frac{-(b-c)}{(a-b)(b-c)(c-a)} - \frac{(c-a)}{(a-b)(b-c)(c-a)} - \frac{(a-b)}{(a-b)(b-c)(c-a)}

Combine the numerators: Numerator =−(b−c)−(c−a)−(a−b)= -(b-c) - (c-a) - (a-b) Numerator =−b+c−c+a−a+b= -b + c - c + a - a + b Numerator =(−b+b)+(c−c)+(a−a)= (-b+b) + (c-c) + (a-a) Numerator =0+0+0=0= 0 + 0 + 0 = 0

So, the sum of the exponents E=0(a−b)(b−c)(c−a)E = \frac{0}{(a-b)(b-c)(c-a)}. Since a,b,ca, b, c are distinct real numbers, none of (a−b)(a-b), (b−c)(b-c), or (c−a)(c-a) are zero. Therefore, the denominator (a−b)(b−c)(c−a)(a-b)(b-c)(c-a) is non-zero. Thus, E=0E = 0.

Alternative method for simplifying the exponent:

Let P=a−bP = a-b, Q=b−cQ = b-c, R=c−aR = c-a.

Notice that P+Q+R=(a−b)+(b−c)+(c−a)=0P+Q+R = (a-b) + (b-c) + (c-a) = 0. Also, a−c=−(c−a)=−Ra-c = -(c-a) = -R. b−a=−(a−b)=−Pb-a = -(a-b) = -P. c−b=−(b−c)=−Qc-b = -(b-c) = -Q.

Substitute these into the exponent EE: E=1P(−R)+1Q(−P)+1R(−Q)E = \frac{1}{P(-R)} + \frac{1}{Q(-P)} + \frac{1}{R(-Q)} E=−1PR+−1QP+−1RQE = \frac{-1}{PR} + \frac{-1}{QP} + \frac{-1}{RQ} E=−(1PR+1QP+1RQ)E = -\left( \frac{1}{PR} + \frac{1}{QP} + \frac{1}{RQ} \right)

To add these fractions, the common denominator is PQRPQR: E=−(QPQR+RPQR+PPQR)E = -\left( \frac{Q}{PQR} + \frac{R}{PQR} + \frac{P}{PQR} \right) E=−P+Q+RPQRE = -\frac{P+Q+R}{PQR}

Since P+Q+R=0P+Q+R=0, we have: E=−0PQR=0E = -\frac{0}{PQR} = 0. (Note: P,Q,RP, Q, R are non-zero because a,b,ca, b, c are distinct.)

Step 4: Substitute the simplified exponent back into the expression.

The original expression simplifies to xE=x0x^E = x^0. Assuming x≠0x \neq 0, x0=1x^0 = 1.

The numerical value of the expression is 1.