\nThe given quadratic equation is x\u00b2 – 7x + 11 = 0.
\nLet the roots be \u03b1 and \u03b2.
\nAccording to Vieta’s formulas, for a quadratic equation ax\u00b2 + bx + c = 0, the sum of the roots is \u03b1 + \u03b2 = -b\/a and the product of the roots is \u03b1\u03b2 = c\/a.
\nFor the given equation x\u00b2 – 7x + 11 = 0 (where a=1, b=-7, c=11):
\nSum of roots: \u03b1 + \u03b2 = -(-7)\/1 = 7.
\nProduct of roots: \u03b1\u03b2 = 11\/1 = 11.
\nWe need to find the value of \u03b1\u00b3 + \u03b2\u00b3.
\nWe can use the algebraic identity for the sum of cubes: \u03b1\u00b3 + \u03b2\u00b3 = (\u03b1 + \u03b2)(\u03b1\u00b2 – \u03b1\u03b2 + \u03b2\u00b2).
\nWe can express \u03b1\u00b2 + \u03b2\u00b2 in terms of (\u03b1 + \u03b2) and \u03b1\u03b2:
\n\u03b1\u00b2 + \u03b2\u00b2 = (\u03b1 + \u03b2)\u00b2 – 2\u03b1\u03b2.
\nSubstituting this into the identity:
\n\u03b1\u00b3 + \u03b2\u00b3 = (\u03b1 + \u03b2)[((\u03b1 + \u03b2)\u00b2 – 2\u03b1\u03b2) – \u03b1\u03b2]
\n\u03b1\u00b3 + \u03b2\u00b3 = (\u03b1 + \u03b2)((\u03b1 + \u03b2)\u00b2 – 3\u03b1\u03b2).
\nNow, substitute the values we found for (\u03b1 + \u03b2) and \u03b1\u03b2:
\n\u03b1 + \u03b2 = 7
\n\u03b1\u03b2 = 11
\n\u03b1\u00b3 + \u03b2\u00b3 = (7)((7)\u00b2 – 3 \u00d7 11)
\n\u03b1\u00b3 + \u03b2\u00b3 = 7(49 – 33)
\n\u03b1\u00b3 + \u03b2\u00b3 = 7(16)
\n\u03b1\u00b3 + \u03b2\u00b3 = 112.
\n<\/section>\n