AMC 12A 2023 (Problem 12)What is the value of 23−13+43−33+63−53+⋯+183−1732^3-1^3+4^3-3^3+6^3-5^3+\cdots+18^3-17^323−13+43−33+63−53+⋯+183−173?(A) 2023\text{(A)}\;2023(A)2023(B) 2679\text{(B)}\;2679(B)2679(C) 2941\text{(C)}\;2941(C)2941(D) 3159\text{(D)}\;3159(D)3159(E) 3235\text{(E)}\;3235(E)3235Related TopicsCoreToolkit 3 — Sum of cubesHints (4)Hint 1Rewrite the expression as−(13+23+⋯+183)+2(23+43+63+⋯+183).-\left(1^3+2^3+\cdots+18^3\right)+2\left(2^3+4^3+6^3+\cdots+18^3\right).−(13+23+⋯+183)+2(23+43+63+⋯+183).Hint 2By Toolkit 3 — Sum of cubes,13+23+⋯+183=(18⋅192)2=(9⋅19)2.1^3+2^3+\cdots+18^3=\left(\frac{18\cdot19}{2}\right)^2=(9\cdot19)^2.13+23+⋯+183=(218⋅19)2=(9⋅19)2.Hint 3By Toolkit 3 — Sum of cubes,2(23+43+63+⋯+183)2\left(2^3+4^3+6^3+\cdots+18^3\right)2(23+43+63+⋯+183)=2⋅23(13+23+33+⋯+93)=2\cdot2^3\left(1^3+2^3+3^3+\cdots+9^3\right)=2⋅23(13+23+33+⋯+93)=16(9⋅102)2=16\left(\frac{9\cdot10}{2}\right)^2=16(29⋅10)2=4⋅902.=4\cdot90^2.=4⋅902.Hint 4By Hints 1, 2, and 3,−(13+23+⋯+183)+2(23+43+63+⋯+183)-\left(1^3+2^3+\cdots+18^3\right)+2\left(2^3+4^3+6^3+\cdots+18^3\right)−(13+23+⋯+183)+2(23+43+63+⋯+183)=−(9⋅19)2+4⋅902=-(9\cdot19)^2+4\cdot90^2=−(9⋅19)2+4⋅902=92(−192+400)=9^2(-19^2+400)=92(−192+400)=81⋅39=3159.=81\cdot39=3159.=81⋅39=3159.Final Answer(D) 315931593159