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Cube Roots
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I have managed to solve it based on assuming x to be an integer value.
Considering cube root integers, the cube root of 8 is 2, 27 is 3 and 64 is 4; if x is an integer it is likely to be one of these numbers. Substituting 8 as x gives a correct solution.
The equation becomes 3√8 + 3√(8-16) = 3√8-8
We get 2 -2 = 0
Of course there is likely to be other non-integer solutions.
I wonder, if in a maths exam is guessed the solution, would I get all or part of the question’s marks.
Considering cube root integers, the cube root of 8 is 2, 27 is 3 and 64 is 4; if x is an integer it is likely to be one of these numbers. Substituting 8 as x gives a correct solution.
The equation becomes 3√8 + 3√(8-16) = 3√8-8
We get 2 -2 = 0
Of course there is likely to be other non-integer solutions.
I wonder, if in a maths exam is guessed the solution, would I get all or part of the question’s marks.
Let y = x - 8 and then cube both sides (this isn't strictly necessary, but makes some of the intermediate working easier because symmetry). The equation becomes (note that I will use cbrt for cube roots throughout):
cbrt(y + 8) + cbrt(y - 8) = cbrt(y)
(y + 8) + (y - 8) + 3 [cbrt((y + 8)(y - 8)^2) + cbrt((y + 8)^2(y - 8))] = y
or, simplifying,
y + 3 cbrt(y^2 - 64) [cbrt(y + 8) + cbrt(y - 8)] = 0
by stealing the starting equation, we can reduce this to just
y + 3 cbrt(y^2 - 64)*cbrt(y) = 0
and then rearrange (taking the cube-root stuff to the right-hand side) and cube again:
y^3 = -27 y (y^2 - 64) .
The first solution is just y = 0 (or x = 8, recovering Hymie's solution). The second possibility is that
y^2 = -27 (y^2 - 64)
=> y^2 = 27*16/7 = (3/7)*144
=> y = ±12√(3/7),
from which x = 8 ±12√(3/7) forms a second pair of solutions.
We should formally check that these also work, since it's often possible to introduce "false" solutions by "undoing" roots or squaring square roots. However, we're saved from having to do this here because the cube root is a bijective function over real numbers (ie, for every real-numbered input there's a unique real-numbered output).
Still, if you check, then you find that the equation does work for these other solutions, with left and right-hand sides being about ±1.9879 for x = 8 ±12√(3/7) respectively.
In the end, if you didn't bother to do the y = x - 8 thing, then the only extra complication would be having to solve the quadratic equation
(x - 8)^2 = -27x(x - 16),
which rearranges to
7x^2 - 112 x + 16 = 0,
with the same solutions as before. But it's much, much easier to go wrong doing it this way, by hand at least, so rearranging to start with makes things a helluva lot easier.
cbrt(y + 8) + cbrt(y - 8) = cbrt(y)
(y + 8) + (y - 8) + 3 [cbrt((y + 8)(y - 8)^2) + cbrt((y + 8)^2(y - 8))] = y
or, simplifying,
y + 3 cbrt(y^2 - 64) [cbrt(y + 8) + cbrt(y - 8)] = 0
by stealing the starting equation, we can reduce this to just
y + 3 cbrt(y^2 - 64)*cbrt(y) = 0
and then rearrange (taking the cube-root stuff to the right-hand side) and cube again:
y^3 = -27 y (y^2 - 64) .
The first solution is just y = 0 (or x = 8, recovering Hymie's solution). The second possibility is that
y^2 = -27 (y^2 - 64)
=> y^2 = 27*16/7 = (3/7)*144
=> y = ±12√(3/7),
from which x = 8 ±12√(3/7) forms a second pair of solutions.
We should formally check that these also work, since it's often possible to introduce "false" solutions by "undoing" roots or squaring square roots. However, we're saved from having to do this here because the cube root is a bijective function over real numbers (ie, for every real-numbered input there's a unique real-numbered output).
Still, if you check, then you find that the equation does work for these other solutions, with left and right-hand sides being about ±1.9879 for x = 8 ±12√(3/7) respectively.
In the end, if you didn't bother to do the y = x - 8 thing, then the only extra complication would be having to solve the quadratic equation
(x - 8)^2 = -27x(x - 16),
which rearranges to
7x^2 - 112 x + 16 = 0,
with the same solutions as before. But it's much, much easier to go wrong doing it this way, by hand at least, so rearranging to start with makes things a helluva lot easier.
TTT \I could solve it \
Provide another computer program and BA is yours!!
This is not to undermine your effort Clare and thank you!! Hopefully you'll agree that to look at problem and convert it into computer code not only requires an understanding of what's going on with the question, but raises the bar to a new level when seeking solutions.
Provide another computer program and BA is yours!!
This is not to undermine your effort Clare and thank you!! Hopefully you'll agree that to look at problem and convert it into computer code not only requires an understanding of what's going on with the question, but raises the bar to a new level when seeking solutions.
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