If a b and y are the roots of 3 x 3 + 8 x 2 − 1 = 0 find ( b + 1 / y ) ( y + 1 / a ) ( a + 1 / b )

Question

If a b and y are the roots of   3      x    3    +  8      x    2    −  1  =  0 find   (  b  +  1      /    y  )  (  y  +  1      /    a  )  (  a  +  1      /    b  )

Sophia Mcdowell · Accepted Answer

You have a few mistakes:  (  b  +      1    y    )  (  y  +      1    a    )  (  a  +      1    b    )  ≠      a    2        b    2        y    2    +      1          a      b      y      , as you write above  (  b  +      1    y    )  (  y  +      1    a    )  (  a  +      1    b    )  ≠                    a        2                    b        2                    y        2            +      1              a      b      y      , as you apply the formula you wroteThe actual solution is:  (  b  +      1    y    )  (  y  +      1    a    )  (  a  +      1    b    )  =      1          a      b      y        (  y  b  +  1  )  (  a  y  +  1  )  (  a  b  +  1  )  =      1          a      b      y        (  a  b      y    2    +  a  y  +  y  b  +  1  )  (  a  b  +  1  )  =      1          a      b      y        (      a    2        b    2        y    2    +      a    2    b  y  +  a      b    2    y  +  a  b      y    2    +  a  b  +  a  y  +  y  b  +  1  )  =  a  b  y  +  a  +  b  +  y  +            a      b      +      a      y      +      y      b              a      b      y        +      1          a      b      y      Now, using Vieta&#039;s Formulas, we have that   a  b  y  =      1    3    ,  a  b  +  a  y  +  y  b  =  0  ,  a  +  b  +  y  =  −      8    3  . This gives   (  b  +      1    y    )  (  y  +      1    a    )  (  a  +      1    b    )  =      1    3    −      8    3    +  0  +  3  =      2    3

Desirae Washington · Accepted Answer

You can also calculate the expression directly using the fact that

p (x) = 3 (x - a) (x - b) (x - y) \Rightarrow 3 a b y = 1

To do so, note that

\begin{array}{rcl} P & = & (b + 1 / y) (y + 1 / a) (a + 1 / b) \\ = & \frac{(b y + 1) (a y + 1) (a b + 1)}{a b y} \\ \overset{a b y = \frac{1}{3}}{=} & 3 (\frac{1}{3 a} + 1) (\frac{1}{3 b} + 1) (\frac{1}{3 y} + 1) \\ \overset{a b y = \frac{1}{3}}{=} & \frac{1}{3} (1 + 3 a) (1 + 3 b) (1 + 3 y) \end{array}

Now, a standard trick uses the observation

\begin{array}{rcl} p (\frac{t - 1}{3}) & = & 3 (\frac{t - 1}{3} - a) (\frac{t - 1}{3} - b) (\frac{t - 1}{3} - y) \\ = & \frac{1}{9} (t - (1 + 3 a)) (t - (1 + 3 b)) (t - (1 + 3 y)) \end{array}

So, you only need the constant member

c = - \frac{1}{9} (1 + 3 a) (1 + 3 b) (1 + 3 y) = - \frac{1}{3} P

of

p (\frac{t - 1}{3}) = 3 {(\frac{t - 1}{3})}^{3} + 8 {(\frac{t - 1}{3})}^{2} - 1

Hence,

P = - 3 c = \frac{2}{3}

If a b and y are the roots of 3 x 3 </msup> + 8 x 2 </msup> &#x22

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