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In isosceles triangle with base $a$, vertical angle $20$ degrees and lateral side each of length $b$, prove that $a^3+b^3=3ab^2$

Can someone draw picture of what is going on in the question? I don't know meaning of vertical angle and lateral sides

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    The vertical angle is the angle opposite the base. The other two angles of an isosceles triangle are equal, so if the vertical angle is $20^\circ$ the other two angles are $80^\circ$ each. The lateral sides are the two equal sides as distinguished from the base. – 2017-02-11

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Draw the following.

$\Delta ABC$, where $\measuredangle ABC=20^{\circ}$, $AC=a$ and $AB=BC=b$.

Let $M\in BC$ such that $AM=a$.

Hence, $\measuredangle BAM=80^{\circ}-20^{\circ}=60^{\circ}.$

Now, let $N\in AB$ such that $AN=a$.

Hence, $AN=AM=MN=a$ and $\measuredangle NMB=180^{\circ}-80^{\circ}-60^{\circ}=40^{\circ}$.

Now, let $K\in BC$ such that $NK=a$.

Thus, $\measuredangle NKM=40^{\circ}$, which says that $\measuredangle BNK=20^{\circ}$ and from here $BK=a$.

Let $KL$ is an altitude of $\Delta BNK$ and $MF$ is an altitude of $\Delta AMC.$

Hence, since $BN=b-a$ and $\Delta BLK\cong\Delta AFM$, we obtain $AF=BL=\frac{b-a}{2}$

and $FC=a-\frac{b-a}{2}=\frac{3a-b}{2}$.

Also we see that $\Delta ABC\thicksim\Delta CAM$, which gives

$\frac{MC}{a}=\frac{b}{a}$ or $MC=\frac{b^2}{a}$.

Since $AM^2-AF^2=MC^2-FC^2$, we obtain $$a^2-\left(\frac{b-a}{2}\right)^2=\left(\frac{a^2}{b}\right)^2-\left(\frac{3a-b}{2}\right)^2$$ or $$a^3+b^3=3ab^2.$$ Done!

Another way.

$$\frac{a}{\sin20^{\circ}}=\frac{b}{\sin80^{\circ}}$$ or $$4\cos20^{\circ}\cos40^{\circ}=\frac{b}{a}$$

But $\cos20^{\circ}=\frac{BL}{BK}=\frac{b-a}{2a}$.

Thus, $$4\cdot\frac{b-a}{2a}\left(2\left(\frac{b-a}{2a}\right)^2-1\right)=\frac{b}{a}$$ or $$a^3+b^3=3ab^2.$$ Done!

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    @Taylor Ted I am ready to show a proof without trigonometry. – 2017-02-11
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    Can we use sine formula to prove this one? Also i would like to see other proofs of this too?thanks – 2017-02-11
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    @Taylor Ted See proof of stud_iisc. He used $\sin3x=3\sin x-4\sin^3x$. – 2017-02-11
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    yes i have seen it. i was saying that can this be proved if io use (sin20)/a=(sin80)/b=(sin80)/b – 2017-02-11
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    @Taylor Ted I added something. Maybe it's which you wished. – 2017-02-11
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Isosceles triangle with vertical axis of symmetry

This is an Isosceles triangle with vertical axis of symmetry with base length $a$, the other 2 equal length sides (lateral sides) and the corresponding length being $b$. The angle seen from the top vertex (not in the base) is the vertical angle, and is given to be $20^\circ$.

So for proving $a^3 + b^3 = 3ab^2$:

Draw a perpendicular from the top vertex. Now we can infer that $\sin(10) = a/2b$. We know that $\sin(30) = 3\sin (10) - 4 \sin^3 (10)$. This implies $$\frac{1}{2} = 3\frac{a}{2b} - 4 \left(\frac{a}{2b}\right)^3. \blacksquare$$

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    Can we use sin formula here to prove – 2017-02-11
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    Yup and Michael Rozenberg has updated his answer, check that out. – 2017-02-11