if a is equal to root 5 + 1 / root 5 minus 1 and b is equal to root 5 - 1 / 5 + 1 then a minus b whole cube divided by a plus b the whole cube equal to how
Answers
Answer:
Hey mate!
_______________________
Given :
x = 9 - 4 \sqrt{5}x=9−4
5
To find :
x {}^{3} + \frac{1}{x {}^{3} }x
3
+
x
3
1
Solution :
\begin{lgathered}x = 9 - 4 \sqrt{5} \\ \\ \frac{1}{x} = \frac{1}{9 - 4 \sqrt{5}} \times \frac{9 + 4 \sqrt{5}}{9 + 4 \sqrt{5}} \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{(9) {}^{2} - (4 \sqrt{5} ) {}^{2} } \\ \\ \frac{1}{x} = \frac{9 + 4 \sqrt{5} }{81 - 80} \\ \\ \frac{1}{x} = 9 + 4 \sqrt{5}\end{lgathered}
x=9−4
5
x
1
=
9−4
5
1
×
9+4
5
9+4
5
x
1
=
(9)
2
−(4
5
)
2
9+4
5
x
1
=
81−80
9+4
5
x
1
=9+4
5
Now,
\begin{lgathered}x + \frac{1}{x} = 9 - \cancel{4 \sqrt{5} } + 9 + \cancel{4 \sqrt{5}} \\ \\ x + \frac{1}{x} = 9 + 9 \\ \\ x + \frac{1}{x} = 18\end{lgathered}
x+
x
1
=9−
4
5
+9+
4
5
x+
x
1
=9+9
x+
x
1
=18
And, on cubing both sides.
\begin{lgathered}(x + \frac{1}{x} ) {}^{3} = (18){}^{3} \\ \\ x{}^{3} + \frac{1}{x{}^{3} } + 3(x + \frac{1}{x} ) = 5826 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } + 3 \times 18 = 5826 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } + 54 = 5826 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } = 5826 - 54 \\ \\ x{}^{3} + \frac{1}{x{}^{3} } = 5778\end{lgathered}
(x+
x
1
)
3
=(18)
3
x
3
+
x
3
1
+3(x+
x
1
)=5826
x
3
+
x
3
1
+3×18=5826
x
3
+
x
3
1
+54=5826
x
3
+
x
3
1
=5826−54
x
3
+
x
3
1
=5778