if a=1/3-2√2 and b=1/3+2√2 evaluate a²b+ab²
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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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Answer:
a²b+ab²
= ab(a+b)
ab = (1/3-2√2)(1/3+2√2) = 1/9-8 = -71/9
a+b = 1/3-2√2+1/3-2√2 = 2/3
Therefore ab(a+b) = -71/9×2/3 = -142/27
Step-by-step explanation:
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