In triangle ABC (fig. (b)) , internal bisectors of <B and <C meet at p and external bisectors of these angles meet at Q. prove that <BPC + <BQC = 180°
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Answers
Answer:
a
3
−
a
3
1
+2=
8
3
Step-by-step explanation:
If
\begin{gathered}2a - \frac{2}{a} + 1 = 0...eq1 \\ \end{gathered}
2a−
a
2
+1=0...eq1
then to find the value of
\begin{gathered} {a}^{3} - \frac{1}{ {a}^{3} } + 2 \\ \\ \end{gathered}
a
3
−
a
3
1
+2
take cube of eq1
\begin{gathered}2a - \frac{2}{a} = - 1 \\ \\ a - \frac{1}{a} = \frac{ - 1}{2} \\ \\ {(a - \frac{1}{a}) }^{3} = {( \frac{ - 1}{2} )}^{3} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3 {a}^{2} ( \frac{1}{a} ) + 3a( \frac{1}{ {a}^{2} } ) = - \frac{1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3a + \frac{3}{a} = \frac{ - 1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3(a - \frac{1}{a}) = \frac{ - 1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } - 3( - \frac{1}{2}) = \frac{ - 1}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } + \frac{3}{2} = \frac{ - 1}{8}\end{gathered}
2a−
a
2
=−1
a−
a
1
=
2
−1
(a−
a
1
)
3
=(
2
−1
)
3
a
3
−
a
3
1
−3a
2
(
a
1
)+3a(
a
2
1
)=−
8
1
a
3
−
a
3
1
−3a+
a
3
=
8
−1
a
3
−
a
3
1
−3(a−
a
1
)=
8
−1
a
3
−
a
3
1
−3(−
2
1
)=
8
−1
a
3
−
a
3
1
+
2
3
=
8
−1
\begin{gathered}{a}^{3} - \frac{1}{ {a}^{3} } + 2 = \frac{ - 1}{8} - \frac{3}{2} + 2 \\ \\ = \frac{ - 1 - 12 + 16}{8} \\ \\ {a}^{3} - \frac{1}{ {a}^{3} } + 2 = \frac{3}{8} \\ \\ \end{gathered}
a
3
−
a
3
1
+2=
8
−1
−
2
3
+2
=
8
−1−12+16
a
3
−
a
3
1
+2=
8
3
Hope it helps you.
Step-by-step explanation:
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