Question : The difference between the cubes of two given natural numbers is 6272, while the positive difference between the two given numbers is 8. What is the sum of the cubes of the two given numbers?
Option 1: 9728
Option 2: 9684
Option 3: 8000
Option 4: 9600
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Correct Answer: 9728
Solution :
Given: $a^{3}–b^{3}=6272$ and $a–b=8$
We know that $(a–b)^{3}=a^{3}–b^{3}–3ab(a–b)$
Putting the given value, we have,
⇒ $(8)^{3}=6272–3ab(8)$
⇒ $24ab=6272–512$
⇒ $ab=\frac{5760}{24}=240$ ------------------------------------(1)
Putting the value of $a=b+8$ we have,
⇒ $(b+8)b=240$
⇒ $b^{2}+8b–240=0$
⇒ $b^{2}+20b–12b–240=0$
⇒ $b(b+20)–12(b+20)=0$
⇒ $(b+20)(b–12)=0$
⇒ $b=12,–20$
The value of a natural number can't be negative.
Putting the value of $b=12$ we have,
$a=20$
Now, we find, $a^{3}+b^{3}=20^{3}+12^{3}$
⇒ $a^{3}+b^{3}=8000+1728$
⇒ $a^{3}+b^{3}=9728$
Hence, the correct answer is 9728.
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