যদি $\sqrt[3]{x + iy} = p + iq$ হয়, তবে দেখাও যে, $4 (p^{2} - q^{2}) = \frac{x}{p} + \frac{y}{q}$

Created: 1 year ago | Updated: 1 year ago

Updated: 1 year ago

$\sqrt[3]{x + iy} = p + iq$

$\Rightarrow x + iy = p^{3} + 3 (iq) p^{2} + 3 p {(iq)}^{2} + i^{3} q^{3}$

$= p^{3} - 3 {pq}^{2} + i (3 p^{2} q - q^{3})$

$∴ x = p^{3} - 3 {pq}^{2}$

$y = 3 p^{2} q - q^{3}$

$∴ R . S = \frac{x}{p} + \frac{y}{q} = \frac{p^{3} - 3 {pq}^{2}}{p} + \frac{3 p^{2} q - p^{3}}{q}$

$= p^{2} - 3 q^{2} + 3 p^{2} - q^{2} = 4 (p^{2} - q^{2}) = L . S (Proved)$

$\sqrt{\frac{q}{p}} = - \sqrt{\frac{q}{p}} + \sqrt{\frac{q}{p}} = 0$

$∴ L . S = R . S (Proved)$

1 year ago

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