AMC 10A 2020 (Problem 14)Real numbers xxx and yyy satisfy x+y=4x + y = 4x+y=4 and x⋅y=−2x \cdot y = -2x⋅y=−2. What is the value of x+x3y2+y3x2+yx + \dfrac{x^3}{y^2} + \dfrac{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 TopicsToolkit 6 — Square of a sumToolkit 12 — Sum of cubes (factored)Hints (7)Hint 1Calculate x2+y2x^2 + y^2x2+y2.Hint 2By using 6. Square of a Sum: x2+y2=(x+y)2−2xy=42−2(−2)=20x^2 + y^2 = (x + y)^2 - 2xy = 4^2 - 2(-2) = 20x2+y2=(x+y)2−2xy=42−2(−2)=20Hint 3Calculate x3+y3x^3 + y^3x3+y3.Hint 4By using 12. Sum of cubes: x3+y3=(x+y)(x2−xy+y2)=4(20+2)=88x^3 + y^3 = (x + y)(x^2 - xy + y^2) = 4(20 + 2) = 88x3+y3=(x+y)(x2−xy+y2)=4(20+2)=88Hint 5Calculate x5+y5x^5 + y^5x5+y5.Hint 6x5+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) But x2y3+x3y2=x2y2(x+y)x^2y^3 + x^3y^2 = x^2y^2(x + y)x2y3+x3y2=x2y2(x+y). Since xy=−2⇒x2y2=4xy = -2 \Rightarrow x^2y^2 = 4xy=−2⇒x2y2=4, and x+y=4x + y = 4x+y=4, we have x2y2(x+y)=4⋅4=16x^2y^2(x + y) = 4 \cdot 4 = 16x2y2(x+y)=4⋅4=16. x5+y5=(20)(88)−16=1760−16=1744x^5 + y^5 = (20)(88) - 16 = 1760 - 16 = 1744x5+y5=(20)(88)−16=1760−16=1744Hint 7We want x+x3y2+y3x2+y=x+y+x5+y5x2y2x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y = x + y + \dfrac{x^5 + y^5}{x^2y^2}x+y2x3+x2y3+y=x+y+x2y2x5+y5 x+y=4x + y = 4x+y=4, x5+y5=1744x^5 + y^5 = 1744x5+y5=1744, x2y2=4x^2y^2 = 4x2y2=4. x5+y5x2y2=17444=436\dfrac{x^5 + y^5}{x^2y^2} = \dfrac{1744}{4} = 436x2y2x5+y5=41744=436 Therefore, x+x3y2+y3x2+y=4+436=440x + \dfrac{x^3}{y^2} + \dfrac{y^3}{x^2} + y = 4 + 436 = 440x+y2x3+x2y3+y=4+436=440Final Answer(D) 440440440