Show that:1. (4pq + 3q)2 – (4pq – 3q)2 = 48pq22. (a – b)(a + b)

1.	Show that:1. (4pq + 3q)2 – (4pq – 3q)2 = 48pq22. (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0
Answer» ANSWER: $1. )\: {(4pq + 3q)}^{2} - {(4pq - 3q)}^{2} = {48pq}^{2} \\ = 16pq + 9q - 16pq - 9q = 2304pq \\ = (16 - 16)pq + (9 - 9)q = 2304pq \\ = 0 + 0 = 2304pq \\ 2304pq$

Show that:1. (4pq + 3q)2 – (4pq – 3q)2 = 48pq22. (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

Answer»

$1. )\: {(4pq + 3q)}^{2} - {(4pq - 3q)}^{2} = {48pq}^{2} \\ = 16pq + 9q - 16pq - 9q = 2304pq \\ = (16 - 16)pq + (9 - 9)q = 2304pq \\ = 0 + 0 = 2304pq \\ 2304pq$

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Show that:1. (4pq + 3q)2 – (4pq – 3q)2 = 48pq22. (a – b)(a + b) + (b – c) (b + c) + (c – a) (c + a) = 0

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