Question

Если a + 1/a = 10, то a^3+a^(-3)Ответы:480970630720

Answer 1

Ответ:

970

Пошаговое объяснение:

(а+1/а)^3=(a^3+1/(a)^3)+3(a*1/a)(a+1/a)=(a^3+1/(a)^3)+3*10

1000-30=(a^3+1/(a)^3)

(a^3+1/(a)^3)=970

Answer 2

a³+a⁻³=(a + (1/a))*(а²-а/а+1/а²)=10*(а²-1+1/а²)

(a + 1/a)²=а²+2а/а+1/а²=100⇒а²+2+1/а²=100; а²+1/а²=100-2=98;

a³+a⁻³=10*(98-1)=970

ОТВЕТ   970

Answer 3

To find the value of $a^3 + a^{-3}$ when $a + \frac{1}{a} = 10$, we can use the following method:

We are given that $a + \frac{1}{a} = 10$. Let's try to manipulate this expression to get $a^3 + a^{-3}$.

First, let's square the given equation: $(a + \frac{1}{a})^2 = 10^2$ $a^2 + 2a\cdot\frac{1}{a} + \frac{1}{a^2} = 100$ $a^2 + 2 + \frac{1}{a^2} = 100$

Next, let's find the value of $a^2 + \frac{1}{a^2}$ using the information given in the original equation. We can square the original equation to get $a^2 + 2a\cdot\frac{1}{a} + \frac{1}{a^2} = 100$, and since $a + \frac{1}{a} = 10$, we have: $a^2 + 2 + \frac{1}{a^2} = 100$

Subtracting 2 from both sides: $a^2 + \frac{1}{a^2} = 98$

Now, let's find $a^3 + a^{-3}$ using the cube of the original equation: $(a + \frac{1}{a})^3 = 10^3$ $a^3 + 3a^2\cdot\frac{1}{a} + 3a\cdot\frac{1}{a^2} + \frac{1}{a^3} = 1000$ $a^3 + 3a + 3\frac{1}{a} + \frac{1}{a^3} = 1000$

We can see that $3a + 3\frac{1}{a} = 3(a + \frac{1}{a})$. And since $a + \frac{1}{a} = 10$, we have: $3a + 3\frac{1}{a} = 3 \times 10 = 30$

Therefore, the equation becomes: $a^3 + 30 + \frac{1}{a^3} = 1000$

Now, to find $a^3 + a^{-3}$, we just need to subtract 30 from both sides: $a^3 + \frac{1}{a^3} = 1000 - 30 = 970$

So the final answer is $a^3 + a^{-3} = 970$