Question

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Answer 1

Answer:

Given

$a + bk^{1/3} + ck^{2/3} = 0\qquad\text{(1)}$

Multiplying equation (1) by b gives

$ab + b^2k^{1/3} + bck^{2/3} = 0\qquad\text{(2)}$

Multiplying equation (1) by $ck^{1/3}$ gives

$c^2k + ack^{1/3} + bck^{2/3} = 0\qquad\text{(3)}$

Subtracting equation (3) from equation (2) gives

$(ab-c^2k) + (b^2-ac)k^{1/3} = 0$

If b²-ac ≠ 0, this would yield

$k^{1/3} = (c^2k-ab) / (b^2-ac)$

which is rational, contrary to k not being a rational cube.

Therefore b² = ac.

Similarly, if ab-c²k ≠ 0, this would yield

$k^{2/3} = (ac-b^2)/(ab-c^2k)$

contrary to k not being a rational cube.

Therefore c²k = ab.

Hence

b³k = (b²)(bk) = (ac)(bk) = (ab)(ck) = (c²k)(ck) = c³k².

If c ≠ 0, this then implies that k = ( b / c )³, contrary to k not being a rational cube.  Therefore c = 0.

From b² = ac, it follows that b = 0.

From equation (1) it now follows that a = 0.

Hence a = b = c = 0.

Answer 2

Answer:

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