Question

Plzzzzzzzzz answer meeeeeee

Answer 1

$x = 2 \\ y = 57 \\ \\ \frac{d^3y}{dx^3} = \frac{57d^3}{8d} \\ \\ = 7\frac{1}{8} \times d^2 = [7 + \frac{1}{8}]d^2 \\ \\= [(2^2 + 2 + 1) + \frac{1}{2^3}]d^2 = [2^2 + 2^1 + 2^0 + 2^{- 3}]d^2 \\... \\... \\ x = 4 \\ y = 1345 \\ \\ \frac{d^3y}{dx^3} = \frac{1345d^3}{64d} \\ \\ = 21\frac{1}{64} \times d^2 = [21 + \frac{1}{64}]d^2 \\ \\ = [(4^2 + 4 + 1) + \frac{1}{4^3}]d^2 = [4^2 + 4^1 + 4^0 + 4^{- 3}]d^2 \\... \\...$

$x = 9 \\ y = 66340 \\ \\ \frac{d^3y}{dx^3} = \frac{66340d^3}{729d} \\ \\ = 91\frac{1}{729} \times d^2 = [91 + \frac{1}{729}]d^2 \\ \\ = [(9^2 + 9 + 1) + \frac{1}{9^3}]d^2 = [9^2 + 9^1 + 9^0 + 9^{- 3}]d^2 \\... \\...$

$\\ \\ \\ x = 13 \\ y = 402052 \\ \\ \frac{d^3y}{dx^3} = \frac{402052d^3}{2197d} \\ \\ = 183\frac{1}{2197} \times d^2 = [183 + \frac{1}{2197}]d^2 \\ \\ = [(13^2 + 13 + 1) + \frac{1}{13^3}]d^2 = [13^2 + 13^1 + 13^0 + 13^{- 3}]d^2 \\... \\... \\ x = 16 \\ y = 1118209 \\ \\ \frac{d^3y}{dx^3} = \frac{1118209d^3}{4096d} \\ \\ = 273\frac{1}{4096} \times d^2 = [273 + \frac{1}{4096}]d^2 \\ \\ = [(16^2 + 16 + 1) + \frac{1}{16^3}]d^2 = [16^2 + 16^1 + 16^0 + 16^{- 3}]d^2 \\... \\...$

$\\ \\ \\ x = 21 \\ y = 4287844 \\ \\ \frac{d^3y}{dx^3} = \frac{4287844d^3}{9261d} \\ \\ = 463\frac{1}{9261} \times d^2 = [463 + \frac{1}{9261}]d^2 \\ \\ = [(21^2 + 21 + 1) + \frac{1}{21^3}]d^2 = [21^2 + 21^1 + 21^0 + 21^{- 3}]d^2 \\... \\... \\ x = 29 \\ y = 21242820 \\ \\ \frac{d^3y}{dx^3} = \frac{21242820d^3}{24389d} \\ \\ = 871\frac{1}{24389} \times d^2 = [871 + \frac{1}{24389}]d^2 \\ \\ = [(29^2 + 29 + 1) + \frac{1}{29^3}]d^2 = [29^2 + 29^1 + 29^0 + 29^{- 3}]d^2$

Here we can find that,

$\frac{d^3y}{dx^3} = [x^2 + x^1 + x^0 + x^{- 3}]d^2 \\ \\ \& \\ \\ y = (x^2 + x^1 + x^0)x^3 + 1 \\ \\ y = x^5 + x^4 + x^3 + x^0$

Consider x = 5

$\frac{d^3y}{dx^3} \\ \\ = [5^2 + 5^1 + 5^0 + 5^{- 3}]d^2 \\ \\ = [25 + 5 + 1 + \frac{1}{125}]d^2 \\ \\ = [31 + \frac{1}{125}]d^2 \\ \\ = 31\frac{1}{125} \times d^2 \\ \\ = \frac{3876d^3}{125d} \\ \\ \\ \\ y = 3876 \\ \\ = 3125 + 625 + 125 + 1 \\ \\ = 5^5 + 5^4 + 5^3 + 5^0 \\ \\ = (5^2 + 5^1 + 5^0)5^3 + 5^0 \\ \\ = (25 + 5 + 1)125 + 1$

Hope this may be helpful.

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