In thé questions WE asked to proove that A3+B3 =-8/27

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Answer

α3+β3=−827\alpha^3 + \beta^3 = -\frac{8}{27}

Step-by-step solution

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To prove that α3+β3=−827\alpha^3 + \beta^3 = -\frac{8}{27}, we use the values of α+β\alpha + \beta and αβ\alpha\beta derived from the original quadratic equation 3x2−4x+2=03x^2 - 4x + 2 = 0.

  1. Recall the sum and product of roots.

    For the equation 3x2−4x+2=03x^2 - 4x + 2 = 0, the roots are α\alpha and β\beta.

    α+β=−−43=43\alpha + \beta = -\frac{-4}{3} = \frac{4}{3} αβ=23\alpha\beta = \frac{2}{3}

  2. Use the identity for the sum of cubes.

    The correct identity for α3+β3\alpha^3 + \beta^3 is:

    α3+β3=(α+β)((α+β)2−3αβ)\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha\beta)

  3. Substitute the values of α+β\alpha + \beta and αβ\alpha\beta into the identity.

    α3+β3=(43)((43)2−3(23))\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\left(\frac{4}{3}\right)^2 - 3\left(\frac{2}{3}\right)\right) α3+β3=(43)(169−63)\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - \frac{6}{3}\right) α3+β3=(43)(169−189)\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - \frac{18}{9}\right) α3+β3=(43)(−29)\alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(-\frac{2}{9}\right) α3+β3=−827\alpha^3 + \beta^3 = -\frac{8}{27}
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