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Question Number 71903 by naka3546 last updated on 21/Oct/19

a x + b y = 3

{a x}^{2} + {b y}^{2} = 7

{a x}^{3} + {b y}^{3} = 16

{a x}^{4} + {b y}^{4} = 42

{a x}^{5} + {b y}^{5} = ?

Answered by mind is power last updated on 21/Oct/19

(x + y) ({ax}^{k} + {by}^{k}) = {ax}^{k + 1} + {by}^{k + 1} + xy ({ax}^{k - 1} + {by}^{k - 1})

\Rightarrow {\begin{cases} (x + y) . 3 = 7 + xy (a + b) . . 1 \\ 7 (x + y) = 16 + xy (3) . . 2 \end{cases}

16 (x + y) = 42 + xy . (7) \dots 3

2 & 3 s = x + y and p = xy \Rightarrow

7 s = 16 + 3 p

16 s = 42 + 7 p

16 (\frac{16 + 3 p}{7}) = 42 + 7 p

\Rightarrow 256 + 48 p = 294 + 49 p

p = - 38 = xy

s = \frac{16 - 114}{7} = \frac{- 98}{7} = - 14 = x + y

(x + y) . ({ax}^{4} + {by}^{4}) = {ax}^{5} + {by}^{5} + xy ({ax}^{3} + {by}^{3})

\Rightarrow - 14 (42) = {ax}^{5} + {by}^{5} - 38 (16)

\Rightarrow {ax}^{5} + {by}^{5} = 38.16 - 14.42 = 20

Commented by MJS last updated on 22/Oct/19

great!

Commented by mind is power last updated on 23/Oct/19

thanx sir

Answered by MJS last updated on 21/Oct/19

solve (1) for y, (2) for b, (3) for a; these are

all linear, then (4) for x (quadratic)

\Rightarrow

x = - 7 \pm \sqrt{87}

y = - 7 \mp \sqrt{87}

a = \frac{49}{76} \pm \frac{457 \sqrt{87}}{6612}

b = \frac{49}{76} \mp \frac{457 \sqrt{87}}{6612}

{a x}^{5} + {b y}^{5} = 20

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