if x is equal to 5 minus under root 21 upon to prove that that bracket x cube + 1 upon x cube bracket closed - 5 bracket open X square + 1 upon x square bracket close plus x=0
Answers
Answer:
Step-by-step explanation:
Answer:
We have i )x = \frac{5-\sqrt{21}}{2}--(1)i)x=
2
5−
21
−−(1)
Now,
\begin{gathered}ii)\frac{1}{x}\\=\frac{1}{\frac{5-\sqrt{21}}{2}}\\=\frac{2}{5-\sqrt{21}}\end{gathered}
ii)
x
1
=
2
5−
21
1
=
5−
21
2
Rationalising the denominator, we get
=\frac{2(5+\sqrt{21})}{[(5-2\sqrt{21})(5+2\sqrt{21})]}
[(5−2
21
)(5+2
21
)]
2(5+
21
)
= \frac{2(5+\sqrt{21})}{[(5^{2}-(2\sqrt{21})^{2}]}
[(5
2
−(2
21
)
2
]
2(5+
21
)
=\frac{2(5+\sqrt{21})}{(25-21)}
(25−21)
2(5+
21
)
=\frac{2(5+\sqrt{21})}{4}
4
2(5+
21
)
=\frac{(5+\sqrt{21})}{2}--(2)
2
(5+
21
)
−−(2)
\begin{gathered} iii ) x + \frac{1}{x}\\=\frac{5-\sqrt{21}}{2}+\frac{(5+\sqrt{21})}{2}\end{gathered}
iii)x+
x
1
=
2
5−
21
+
2
(5+
21
)
= \frac{5-\sqrt{21}+5+\sqrt{21}}{2}
2
5−
21
+5+
21
=\frac{10}{2}
2
10
=55 ----(3)
Now ,
\begin{gathered}LHS= \big(x^{3}+\frac{1}{x^{3}}\big)-5\big(x^{2}+\frac{1}{x^{2}}\big)+\big(x+\frac{1}{x}\big)\\=[\big(x+\frac{1}{x}\big)^{3}-3(x+\frac{1}{x})]-5[\big(x+\frac{1}{x}\big)^{2}-2]+(x+\frac{1}{x})\\=[\big(x+\frac{1}{x}\big)^{3}-2(x+\frac{1}{x})-5[\big(x+\frac{1}{x}\big)^{2}-2]\\=5^{3}-2\times 5-5[5^{2}-2]\\=125-10-5(25-2)\\=115-5\times23\\=115-115\\=0\\=RHS\end{gathered}
LHS=(x
3
+
x
3
1
)−5(x
2
+
x
2
1
)+(x+
x
1
)
=[(x+
x
1
)
3
−3(x+
x
1
)]−5[(x+
x
1
)
2
−2]+(x+
x
1
)
=[(x+
x
1
)
3
−2(x+
x
1
)−5[(x+
x
1
)
2
−2]
=5
3
−2×5−5[5
2
−2]
=125−10−5(25−2)
=115−5×23
=115−115
=0
=RHS
Therefore,
\begin{gathered} \big(x^{3}+\frac{1}{x^{3}}\big)-5\big(x^{2}+\frac{1}{x^{2}}\big)+\big(x+\frac{1}{x}\big)\\=0\end{gathered}
(x
3
+
x
3
1
)−5(x
2
+
x
2
1
)+(x+
x
1
)
=0