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Power of a Power

Question 5iii...

Question

Question 5(iii)
Show that:
${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$

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Solution

L.H.S= ${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = {(\frac{4}{3} m)}^{2} - 2 \times \frac{4}{3} m \times \frac{3}{4} n + {(\frac{3}{4} n)}^{2} + 2 m n$
$[U s i n g i d e n t i t y (a - b)^{2} = a^{2} - 2 a b + b^{2}]$
$= \frac{16}{9} m^{2} - 2 m n + \frac{9}{16} n^{2} + 2 m n$
$= \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$

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Question 5(iii)
Show that:
${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$

{(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

{(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

{(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}

Show that

(i) (3x + 7)² − 84x = (3x − 7)²(ii) (9p − 5q)² + 180pq = (9p + 5q)²

(iii)

(iv) (4pq + 3q)²− (4pq − 3q)² = 48pq²

(v) (a − b) (a + b) + (b − c) (b + c) + (c − a) (c + a) = 0

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Power of a Power

Standard VIII Mathematics

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