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(a+ b)3 = a3 + 3a2b + 3ab2 + b3 = (a3 + 3ab2) + (b3 + 3a2b) = 2006 + 2005 = 4011
=> a + b = \(\sqrt[3]{4011}\)
(a - b)3 = a3 - 3a2b + 3ab2 - b3 = (a3 + 3ab2) - (b3 + 3a2b) = 2006 - 2005 = 1
=> a - b = 1
=> P = a2 - b2 = (a - b)(a + b) = \(\sqrt[3]{4011}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: (a3 + 3ab2)2 = a6 + 6a4b2 + 9a2b4 = 20062
(b3 + 3a2b)2 = b6 + 6a2b4 + 9a4b2 = 20052
=> (a3 + 3ab2)2 - (b3 + 3a2b)2 = a6 - 3a4b + 3a2b4 - b6 = 20062 - 20052
Hay (a2 - b2)3 = 4011. Vậy P = a2 - b2 = \(\sqrt[3]{4011}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a^3-3ab^2=19\Rightarrow\left(a^3-3ab^2\right)^2=361\)
\(\Leftrightarrow a^6-6a^4b^2+9a^2b^4=361\left(1\right)\)
\(b^3-3a^2b=98\Rightarrow\left(b^3-3a^2b\right)^2=9604\)
\(\Leftrightarrow b^6-6a^2b^4+9a^4b^2=9604\left(2\right)\)
\(\text{Công 2 vế (1) và (2) ta được :}\)
\(a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=9956\)
\(\Leftrightarrow a^6+3a^4b^2+3a^2b^4+b^6=9956\)
\(\Leftrightarrow\left(a^2+b^2\right)^3=9956\)
\(\Leftrightarrow a^2+b^2=\sqrt[3]{9956}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Có a3-3ab2=10=>(a3-3ab2)2=100(1)
Có b3-3a2b=5=>(b3-3a2b)2=25(2)
Cộng (1) và (2)
=>(a3-3ab2)2+(b3-3a2b)2=100+25
<=>a6-6a4b2+9a2b4+b6-6a2b4+9a2b4=125
<=>a6+3a2b4+3a4b2+b6=125
<=>(a2+b2)3=125
<=>a2+b2=5
vậy a2+b2=5
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Theo bài ra ta có :
\(\left(a^3-3ab^2\right)^2+\left(b^3-3a^2b\right)^2\)
\(=233^2+2010^2\)
\(\Rightarrow\left(a^2+b^2\right)^3=4094389\)
\(\Rightarrow a^2+b^2=\sqrt[3]{4094389}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
TA có :
\(\left(a^3-3ab^2\right)^2+\left(b^3-3a^2b\right)^2=19^2+18^2=685\)
=> \(a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=685\)
=> \(b^6+3a^2b^4+3a^4b^2+b^6=685\Rightarrow\left(a^2+b^2\right)^3=685\)
=> P = ( số hơi lẻ )
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\left\{{}\begin{matrix}a^3-3ab^2=19\\b^3-3a^2b=98\end{matrix}\right.\) => \(\left\{{}\begin{matrix}\left(a^3-3ab^2\right)^2=19^2=361\\\left(b^3-3a^2b\right)^2=98^2=9604\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}a^6-6a^4b^2+9a^2b^4=361\\b^6-6a^2b^4+9a^4b^2=9604\end{matrix}\right.\)
=> \(a^6+b^6+\left(9a^2b^4-6a^2b^4\right)+\left(9b^2a^4-6a^4b^2\right)=9965\)
=> \(a^6+3a^2b^4+3a^4b^2+b^6=9965\)
=> \(\left(a^2+b^2\right)^3=9965\)
=> \(a^2+b^2=\sqrt[3]{9965}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Cách 1:
Ta có: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=\left(a^3+3ab^2\right)+\left(b^3+3a^2b\right)=4011\)
\(\Rightarrow a+b=\sqrt[3]{4011}\)
Mặt khác: \(\left(a-b\right)^3=a^3-3a^2b+3ab^2-b^3=\left(a^3+3ab^2\right)-\left(b^3+3a^2b\right)=1\)
\(\Rightarrow a-b=1\)
Vậy \(a^2-b^2=\left(a+b\right)\left(a-b\right)=\sqrt[3]{4011}.1=\sqrt[3]{4011}\)
Cách 2:
Ta có: \(\left(a^3+3ab^2\right)^2=a^6+6a^4b^2+9a^2b^4\Rightarrow a^6+6a^4b^2+9a^2b^4=2006^2\left(1\right)\)
\(\left(b^3+3a^2b\right)^2=b^6+6a^2b^4+9a^4b^2\Rightarrow b^6+6a^2b^4+9a^4b^2=2005^2\left(2\right)\)
\(\left(1\right)\left(2\right)\Rightarrow\left(a^6+6a^4b^2+9a^2b^4\right)-\left(b^6+6a^2b^4+9a^4b^2\right)=2006^2-2005^2=4011\)
\(\Rightarrow a^6-3a^4b^2+3a^2b^4-b^3=4011\Rightarrow\left(a^2-b^2\right)^3=4011\Rightarrow a^2-b^2=\sqrt[3]{4011}\)
Ta có:a3+3ab2=2006
Và:b3+3a2b=2005
Cộng 2 biểu thức vế với vế ta được:
a3+3ab2+b3+3a2b=2006+2005
=>(a+b)3=4011
=>\(a+b=\sqrt{4011}.\)
Lấy biểu thức thứ nhất trừ biểu thức thứ hai ta dc:
a3+3ab2-b3-3a2b=2006-2005
=>(a-b)3=1
=>a-b=1.
Ta có:\(a^2-b^2=\left(a+b\right)\cdot\left(a-b\right)=\sqrt{4011}\cdot1=\sqrt{4011}.\)
Vậy \(a^2-b^2=\sqrt{4011}.\)