Show that (97)3 + (14)3 is divisible by 111:

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Show that (97)3 + (14)3 is divisible by 111

Solution:

From the question,
(97)3 + (14)3
We know that, a3 + b3 = (a + b) (a2 – ab + b2)
So, (97 + 14) [(97)2 – (97 × 14) + (14)2]
111 [(97)2 – (97 × 14) + (14)2]
Therefore, it is clear that the given expression is divisible by 111.

If a + b = 8 and ab = 15, find the value of a4 + a2b2 + b4.

Solution:

a4 + a2b2 + b4
Above terms can be written as,
a4 + 2a2b2 + b4 – a2b2
(a2)2 + 2a2b2 + (b2)2 – (ab)2
(a2 + b2)2 – (ab)2
(a2 + b2 + ab) (a2 + b – ab)
a + b = 8, ab = 15
So, (a + b)2 = 82
a2 + 2ab + b2 = 64
a2 + 2(15) + b2 = 64
a2 + b2 + 30 = 64
By transposing,
a2 + b2 = 64 – 30
a2 + b2 = 34
Then, a4 + a2b2 + b4
= (a2 + b2 + ab) (a2 + b2 – ab)
= (34 + 15) (34 – 15)
= 49 × 19
= 931

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