1.

If a=1/3-2√2 and b=1/3+2√2 evaluate a²b+ab²​

Answer»

If a = 1/[3–2*2^0.5] and B = 1/[3+2*2^0.5], what is the VALUE of a^2b + AB^2?

a^2b + ab^2 = ab(a+b).

ab = 1/[3–2*2^0.5]*1/[3+2*2^0.5] = 1/(9–8) = 1/1 = 1

a+b = 1/[3–2*2^0.5] + 1/[3+2*2^0.5] = [3+2*2^0.5 +3–2*2^0.5]/1 = 6.

Hence, a^2b + ab^2 = ab(a+b) = 6.

Check: a = 5.828427125, b = 0.171572875.

a^2b + ab^2 = 5.828427125^2*0.171572875 + 5.828427125*0.171572875^2

= 5.828427125 + 0.171572875 = 6. Correct.


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