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MCQMediumJEE 2023Nature of Roots & Formation of Equations

JEE Mathematics 2023 Question with Solution

Let α,β,γ\alpha, \beta, \gamma be the three roots of the equation x3+bx+c=0x^3 + bx + c = 0. If βγ=1=−α\beta\gamma = 1 = -\alpha, then b3+2c3−3α3−6β3−8γ3b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3 is equal to:

  • A

    1558\frac{155}{8}

  • B

    2121

  • C

    1919

  • D

    1698\frac{169}{8}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The roots of x3+bx+c=0x^3 + bx + c = 0 are α,β,γ\alpha, \beta, \gamma with βγ=1=−α\beta\gamma = 1 = -\alpha, so α=−1\alpha = -1 and βγ=1\beta\gamma = 1.

Find: The value of b3+2c3−3α3−6β3−8γ3b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3.

From Vieta’s formulas,

α+β+γ=0,αβ+βγ+γα=b,αβγ=−c.\alpha + \beta + \gamma = 0, \quad \alpha\beta + \beta\gamma + \gamma\alpha = b, \quad \alpha\beta\gamma = -c.

Substituting α=−1\alpha = -1 and βγ=1\beta\gamma = 1,

−1+β+γ=0  ⟹  β+γ=1.-1 + \beta + \gamma = 0 \implies \beta + \gamma = 1.

Also,

(−1)(βγ)=−c  ⟹  −1=−c  ⟹  c=1.(-1)(\beta\gamma) = -c \implies -1 = -c \implies c = 1.

Now,

αβ+βγ+γα=b\alpha\beta + \beta\gamma + \gamma\alpha = b

gives

(−1)β+1+γ(−1)=b  ⟹  −β−γ+1=b.(-1)\beta + 1 + \gamma(-1) = b \implies -\beta - \gamma + 1 = b.

Using β+γ=1\beta + \gamma = 1,

b=−1+1=0.b = -1 + 1 = 0.

So the required expression becomes

03+2(1)3−3(−1)3−6β3−8γ3=5−6β3−8γ3.0^3 + 2(1)^3 - 3(-1)^3 - 6\beta^3 - 8\gamma^3 = 5 - 6\beta^3 - 8\gamma^3.

Since β+γ=1\beta + \gamma = 1 and βγ=1\beta\gamma = 1, β\beta and γ\gamma satisfy

t2−t+1=0.t^2 - t + 1 = 0.

Hence they are −ω-\omega and −ω2-\omega^2, where ω\omega is a cube root of unity.

Therefore,

β3=(−ω)3=−1,γ3=(−ω2)3=−1.\beta^3 = (-\omega)^3 = -1, \quad \gamma^3 = (-\omega^2)^3 = -1.

Substitute into the expression:

5−6(−1)−8(−1)=5+6+8=19.5 - 6(-1) - 8(-1) = 5 + 6 + 8 = 19.

Therefore, the value of the expression is 1919. The correct option is C.

Using the quadratic satisfied by the other two roots

Given: α=−1\alpha = -1, βγ=1\beta\gamma = 1, and α+β+γ=0\alpha + \beta + \gamma = 0.

Find: The value of b3+2c3−3α3−6β3−8γ3b^3 + 2c^3 - 3\alpha^3 - 6\beta^3 - 8\gamma^3.

From α+β+γ=0\alpha + \beta + \gamma = 0,

−1+β+γ=0  ⟹  β+γ=1.-1 + \beta + \gamma = 0 \implies \beta + \gamma = 1.

Also from αβγ=−c\alpha\beta\gamma = -c,

(−1)(1)=−c  ⟹  c=1.(-1)(1) = -c \implies c = 1.

And from αβ+βγ+γα=b\alpha\beta + \beta\gamma + \gamma\alpha = b,

−β+1−γ=b  ⟹  1−(β+γ)=b=0.-\beta + 1 - \gamma = b \implies 1 - (\beta + \gamma) = b = 0.

Thus the expression reduces to

0+2−3(−1)3−6β3−8γ3=5−6β3−8γ3.0 + 2 - 3(-1)^3 - 6\beta^3 - 8\gamma^3 = 5 - 6\beta^3 - 8\gamma^3.

Now use the relations β+γ=1\beta + \gamma = 1 and βγ=1\beta\gamma = 1. The numbers β\beta and γ\gamma are roots of

t2−(β+γ)t+βγ=0  ⟹  t2−t+1=0.t^2 - (\beta + \gamma)t + \beta\gamma = 0 \implies t^2 - t + 1 = 0.

For any root tt of t2−t+1=0t^2 - t + 1 = 0,

t2=t−1.t^2 = t - 1.

Multiplying by tt,

t3=t2−t=(t−1)−t=−1.t^3 = t^2 - t = (t - 1) - t = -1.

Hence,

β3=−1,γ3=−1.\beta^3 = -1, \quad \gamma^3 = -1.

Therefore,

5−6(−1)−8(−1)=19.5 - 6(-1) - 8(-1) = 19.

So the value is 1919.

Common mistakes

  • Using the wrong Vieta relation for the coefficient of x2x^2. Since the equation is x3+bx+c=0x^3 + bx + c = 0, the coefficient of x2x^2 is 00, so α+β+γ=0\alpha + \beta + \gamma = 0. Do not introduce an extra term.

  • Missing the condition 1=−α1 = -\alpha. This means α=−1\alpha = -1, not α=1\alpha = 1. Using the wrong sign changes both cc and the final expression.

  • Stopping after finding β+γ=1\beta + \gamma = 1 and βγ=1\beta\gamma = 1 without extracting β3\beta^3 and γ3\gamma^3. The correct next step is to form t2−t+1=0t^2 - t + 1 = 0 and use it to deduce t3=−1t^3 = -1.

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