The third proportional of a and \(\large\frac{{{b^4}}}{{4a}}\) is:
Solution: Third prop of a and \(\tfrac{b^4}{4a}\) is \(\dfrac{(b^4/4a)^2}{a} = \dfrac{b^8}{16a^3}\).
∴ \(\dfrac{b^8}{16a^3}\)
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The third proportional to (x2 - y2) and (x - y) is:
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