Ratio, Proportion, Indices, LogarithmsPYQ Jan 26Question 4552 of 220
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Find the value of (xbxc)(b+c−a)×(xcxa)(c+a−b)×(xaxb)(a+b−c)\displaystyle \left(\frac{x^b}{x^c}\right)^{(b+c-a)} \times \left(\frac{x^c}{x^a}\right)^{(c+a-b)} \times \left(\frac{x^a}{x^b}\right)^{(a+b-c)}

Options

Axabc\displaystyle x^{abc}
B1
C0
D-1
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Correct Answer

✅ Option b — 1

All Options:

  • Axabc\displaystyle x^{abc}
  • B1
  • C0
  • D-1

Detailed Solution & Explanation

We are given the expression: (xbxc)(b+c−a)×(xcxa)(c+a−b)×(xaxb)(a+b−c)\left(\frac{x^b}{x^c}\right)^{(b+c-a)} \times \left(\frac{x^c}{x^a}\right)^{(c+a-b)} \times \left(\frac{x^a}{x^b}\right)^{(a+b-c)}
Using the division rule of indices, xpxq=xp−q\displaystyle \frac{x^p}{x^q} = x^{p-q}, we can write the expression as: (xb−c)(b+c−a)×(xc−a)(c+a−b)×(xa−b)(a+b−c)(x^{b-c})^{(b+c-a)} \times (x^{c-a})^{(c+a-b)} \times (x^{a-b})^{(a+b-c)}
Using the power rule, (xu)v=xuv\displaystyle (x^u)^v = x^{uv}, the expression becomes: x(b−c)(b+c−a)×x(c−a)(c+a−b)×x(a−b)(a+b−c)x^{(b-c)(b+c-a)} \times x^{(c-a)(c+a-b)} \times x^{(a-b)(a+b-c)}
Using the multiplication rule, xu×xv×xw=xu+v+w\displaystyle x^u \times x^v \times x^w = x^{u+v+w}, we add the exponents: x(b−c)(b+c−a)+(c−a)(c+a−b)+(a−b)(a+b−c)x^{(b-c)(b+c-a) + (c-a)(c+a-b) + (a-b)(a+b-c)}
Let us expand the terms in the exponent: (b−c)(b+c−a)=(b2−c2)−a(b−c)=b2−c2−ab+ac(b-c)(b+c-a) = (b^2 - c^2) - a(b-c) = b^2 - c^2 - ab + ac (c−a)(c+a−b)=(c2−a2)−b(c−a)=c2−a2−bc+ab(c-a)(c+a-b) = (c^2 - a^2) - b(c-a) = c^2 - a^2 - bc + ab (a−b)(a+b−c)=(a2−b2)−c(a−b)=a2−b2−ac+bc(a-b)(a+b-c) = (a^2 - b^2) - c(a-b) = a^2 - b^2 - ac + bc
Adding these three expressions: Exponent=(b2−c2−ab+ac)+(c2−a2−bc+ab)+(a2−b2−ac+bc)\text{Exponent} = (b^2 - c^2 - ab + ac) + (c^2 - a^2 - bc + ab) + (a^2 - b^2 - ac + bc) Exponent=(b2−b2)+(c2−c2)+(a2−a2)+(−ab+ab)+(ac−ac)+(−bc+bc)=0\text{Exponent} = (b^2 - b^2) + (c^2 - c^2) + (a^2 - a^2) + (-ab + ab) + (ac - ac) + (-bc + bc) = 0
Thus, the value of the expression is: x0=1x^0 = 1 Hence, **Option B** is the correct answer.

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