AMC 10A 2020 (Problem 14)Real numbers xxx and yyy satisfy x+y=4x+y=4x+y=4 and xy=−2xy=-2xy=−2. What is the value of x+x3y2+y3x2+yx+\frac{x^3}{y^2}+\frac{y^3}{x^2}+yx+y2x3+x2y3+y?(A) 360\text{(A)}\;360(A)360(B) 400\text{(B)}\;400(B)400(C) 420\text{(C)}\;420(C)420(D) 440\text{(D)}\;440(D)440(E) 480\text{(E)}\;480(E)480Related TopicsCoreToolkit 6 — Square of a sumToolkit 12 — Sum of cubes (factored)Hints (7)Hint 1Calculate x2+y2x^2+y^2x2+y2.Hint 2x2+y2=(x+y)2−2xyx^2+y^2=(x+y)^2-2xyx2+y2=(x+y)2−2xy=42−2(−2)=20=4^2-2(-2)=20=42−2(−2)=20Hint 3Calculate x3+y3x^3+y^3x3+y3.Hint 4By Hint 2,x3+y3=(x+y)(x2−xy+y2)x^3+y^3=(x+y)(x^2-xy+y^2)x3+y3=(x+y)(x2−xy+y2)=4(20−(−2))=88=4\left(20-(-2)\right)=88=4(20−(−2))=88Hint 5Calculate x5+y5x^5+y^5x5+y5.Hint 6By Hints 2 and 4,x5+y5=(x2+y2)(x3+y3)−(x2y3+x3y2)x^5+y^5=(x^2+y^2)(x^3+y^3)-(x^2y^3+x^3y^2)x5+y5=(x2+y2)(x3+y3)−(x2y3+x3y2)=(20)(88)−x2y2(x+y)=(20)(88)-x^2y^2(x+y)=(20)(88)−x2y2(x+y)=1760−(−2)2(4)=1760-(-2)^2(4)=1760−(−2)2(4)=1760−16=1744=1760-16=1744=1760−16=1744Hint 7By Hint 6,x+x3y2+y3x2+yx+\frac{x^3}{y^2}+\frac{y^3}{x^2}+yx+y2x3+x2y3+y=x+y+x5+y5x2y2=x+y+\frac{x^5+y^5}{x^2y^2}=x+y+x2y2x5+y5=4+1744(−2)2=4+\frac{1744}{(-2)^2}=4+(−2)21744=4+17444=4+436=440=4+\frac{1744}{4}=4+436=440=4+41744=4+436=440Final Answer(D) 440440440Related Problems (4)AMC 10B Spring 2021 (Problem 15)AMC 10A 2012 (Problem 22)AMC 10B 2023 (Problem 14)BMO1 2016/2017 (Problem 3)