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Polynomials — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsPolynomials1 Mark
MCQ
If $\text{a}^{\frac{1}{3}}+\text{b}^{\frac{1}{3}}+\text{c}^{\frac{1}{3}}= 0,$ then:
  • A
    a + b + c = 0
  • B
    (a + b + c)3 =27abc
  • C
    a + b + c = 3abc
  • D
    a3 + b3 + c3 = 0
✓

Answer

  1. (a + b + c)3 =27abc
    Solution:
    Let $\text{a}^{\frac{1}{3}}=\text{A},\ \text{b}^{\frac{1}{3}}=\text{B}$ and $\text{c}^{\frac{1}{3}}=\text{C}$
    Now, A + B + C = 0 (given)
    If A + B + C = 0, then A3 + B3 + C3 - 3ABC = 0
    ⇒ A3 + B3 + C3 - 3ABC = 0
    ⇒ A3 + B3 + C3 = 3ABC ...(1)
    $\begin{Bmatrix}\text{A}=\text{a}^{\frac{1}{3}},\ \text{B}=\text{b}^{\frac{1}{3}},\ \text{C}=\text{c}^{\frac{1}{3}}\\\text{A}^3=\text{a},\ \text{B}^3=\text{b},\ \text{C}^3=\text{c}\end{Bmatrix}$
    Then, equation (1) becomes
    $\text{a}+\text{b}+\text{c}=3(\text{abc})^{\frac{1}{3}}$
    Cubing both Sides of above equation, we get
    (a + b + c)3 = 27abc
    Hence, correct option is (b).

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