**If a + b = 8 and ab = 15, find the value of a**^{4} + a^{2}b^{2} + b^{4}.

^{4}+ a

^{2}b

^{2}+ b

^{4}.

**Answer :**

a^{4} + a^{2}b^{2} + b^{4}

Above terms can be written as,

a^{4} + 2a^{2}b^{2} + b^{4} – a^{2}b^{2}

(a^{2})^{2} + 2a^{2}b^{2} + (b^{2})^{2} – (ab)^{2}

(a^{2} + b^{2})^{2} – (ab)^{2}

(a^{2} + b^{2} + ab) (a^{2} + b – ab)

a + b = 8, ab = 15

So, (a + b)^{2} = 8^{2}

a^{2} + 2ab + b^{2} = 64

a^{2} + 2(15) + b^{2} = 64

a^{2} + b^{2} + 30 = 64

By transposing,

a^{2} + b^{2} = 64 – 30

a^{2} + b^{2} = 34

Then, a^{4} + a^{2}b^{2} + b^{4}

= (a^{2} + b^{2} + ab) (a^{2} + b^{2} – ab)

= (34 + 15) (34 – 15)

= 49 × 19

= 931