7. 7.1. The line $y=M$ intersects the graph of the function $y=x^{3}-84 x$ at points with abscissas $a$, $b$, and $c$ ($a<b<c$, see the figure). It turned out that the distance between $a$ and $b$ is half the distance between $b$ and $c$. Find $M$.

![](https://cdn.mathpix.com/cropped/2024_05_06_8a1d5b347fbc679ac0eeg-6.jpg?height=819&width=919&top_left_y=1214&top_left_x=641)
Answer: 160.

## Solution.

1st method. Since the coordinates of the intersection points $a, b$, and $c$ satisfy the equation $x^{3}-84 x-M=0$, by Vieta's theorem, we have the following equalities: $a+b+c=0$ (1), $ab+bc+ac=-84$ (2), $abc=M$ (3). According to the condition, $2(b-a)=c-b$ or $3b=c+2a$. Considering (1), $c=-5b, a=4b$. Substituting into (2): $-5b^{2}+4b^{2}-20b^{2}=-84$, from which we find $b^{2}=4$. Considering that $a$ and $b$ are negative, $b=-2, a=-8, c=10$. Substituting into (3) $M=(-2) \cdot(-8) \cdot 10=160$.

2nd method. Substituting $a, b, c$ for $x$ in the cubic polynomial equation, we get $a^{3}-84a=M, b^{3}-84b=M, c^{3}-84c=M(1)$. Subtracting the first equation from the second, the second from the third, and the third from the first, and dividing the resulting equations by $b-a, c-b$, and $a-c$, respectively, we obtain the following equalities: $a^{2}+ab+b^{2}=84, b^{2}+bc+c^{2}=84, a^{2}+ac+c^{2}=84$. Subtracting the first from the second, we get $b^{2}-a^{2}+b(c-a)+c^{2}-b^{2}=0$. After combining like terms and dividing by $c-a$, we obtain the equality $a+b+c=0$ (2).

According to the condition, $2(b-a)=c-b$ or $3b=c+2a$. Considering (2), $a=4b, c=-5b$.

Adding the equations (1) $a^{3}+b^{3}+c^{3}-84(a+b+c)=3M$, and considering (2), we get $a^{3}+b^{3}+c^{3}=3M$. Substituting the previously obtained values for $a$ and $c$, we then have $-125b^{3}+b^{3}+64b^{3}=3M$, from which $M=-20b^{3}$. Substituting the found $M$ into expression (1): $b^{3}-84b=-20b^{3}$. Solving this equation for $b$, considering that $b<0$, we get $b=-2$. Then $M=160$.