The difference of two consecutive cubes
Correct Answer: |
B) is never divisible by 2 |
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Description for Correct answer:
Difference of two consecutive cubes = \( \Large (x + 1)^{3} - x^{3} \)
\( \Large = x^{3} + 3x^{2} + 3x+ 1 - x^{3} \)
\( \Large = 3x^{2} + 3x + 1 \)
For any value of x, \( \Large 3x^{2} + 3x + 1 \) will always be odd
i.e. never divisible by 2
Checking with examples
For x = 1,
\( \Large = 3x^{2} + 3x + 1 \)
\( \Large = 3(1)^{2} + 3 (1) + 1 \)
= 3 + 3 + 1 = 7 (not divisible by 2)
For x = 2,
\( \Large = 3x^{2} + 3x + 1 \)
\( \Large = 3(2)^{2} + 3(2) + 1 \)
= 12 + 6 + 1
= 19 (not divisible by 2) Hence option (B)is correct.
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Exams >> CDSE >> CDS Maths(2)