tsquare÷3×tcube=??????
Answers
Answered by
2
Step-by-step explanation:
=> t² – 4t + 1 =0
=> t² + 1 = 4t
=> ( t² + 1 ) ÷ t = 4
\begin{gathered}= > \frac{ {t}^{2} }{t} + \frac{1}{t} = 4 \\ \\ = > t + \frac{1}{t} = 4\end{gathered}
=>
t
t
2
+
t
1
=4
=>t+
t
1
=4
Now, Cubing on both sides,
\begin{gathered}= > {(t + \frac{1}{t}) }^{3} = {4 }^{3} \\ \\ \\ = > {t}^{3} + \frac{1}{t ^{3} } + 3(t + \frac{1}{t} )(t \times \frac{1}{t} ) = 64 \\ \\ \\ = > {t}^{3} + \frac{1}{ {t}^{3} } + 3(4)(1) = 64 \\ \\ \\ = > {t}^{3} + \frac{1}{ {t}^{3} } = 64 - 12 \\ \\ \\ = > {t}^{3} + \frac{1}{ {t}^{3} } = 52\end{gathered}
=>(t+
t
1
)
3
=4
3
=>t
3
+
t
3
1
+3(t+
t
1
)(t×
t
1
)=64
=>t
3
+
t
3
1
+3(4)(1)=64
=>t
3
+
t
3
1
=64−12
=>t
3
+
t
3
1
=52
Answered by
0
Answer:
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