ta có : \(\left(sin\alpha+cos\alpha\right)^2=1+2sin\alpha.cos\alpha=\dfrac{49}{25}\)
\(\Rightarrow sin\alpha+cos\alpha=\pm\dfrac{7}{5}\)
ta có : \(A=sin^3\alpha+cos^3\alpha=\left(sin\alpha+cos\alpha\right)^3-3sin\alpha.cos\alpha\left(sin\alpha+cos\alpha\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}A=\left(\dfrac{7}{5}\right)^3-3\left(0,48\right)\left(\dfrac{7}{5}\right)=\dfrac{91}{125}\\A=\left(\dfrac{-7}{5}\right)^3-3\left(0,48\right)\left(\dfrac{-7}{5}\right)=\dfrac{-91}{125}\end{matrix}\right.\)
vậy \(sin^3\alpha+cos^3\alpha=\pm\dfrac{91}{125}\)
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