Show that-i 3x-72-84x-3x-72ii 9p-5q2-180pq - 9p-5q2iii 43m- 34n2 -2mn-169m2- 916n2iv 4pq-3q2-4pq-3q2-48pq2v a-ba-b-b-cb-c-c-ac-a - 0

We know that (a± b)^2 = a^2 + b^2 ± 2ab and, (a b)(a+b) = a^2 b^2 Now,(i) #160; (3x+7)^2 #160; 84x = (3x)^2 + 7^2 + 2(3x)(7) 84x ...

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Byju's Answer

Standard VIII

Mathematics

Algebraic Identities

Show that:i ...

Question

Show that:
(i) $(3 x + 7)^{2} - 84 x = (3 x - 7)^{2}$
(ii) $(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}$
(iii) ${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n = \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$
(iv) $(4 p q + 3 q)^{2} - (4 p q - 3 q)^{2} = 48 p q^{2}$
(v) $(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$

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Solution

We know that
$(a \pm b)^{2} = a^{2} + b^{2} \pm 2 a b$
and,
$(a - b) (a + b) = a^{2} - b^{2}$

Now,
(i)
$(3 x + 7)^{2} - 84 x$
$= (3 x)^{2} + 7^{2} + 2 (3 x) (7) - 84 x$
$= (3 x)^{2} + 7^{2} - 42 x$
$= (3 x - 7)^{2}$

(ii)
$(9 p - 5 q)^{2} + 180 p q$
$= (9 p)^{2} + (5 q)^{2} - 2 (9 p) (5 q) + 180 p q$
$= (9 p)^{2} + (5 q)^{2} - 90 p q + 180 p q$
$= (9 p)^{2} + (5 q)^{2} + 90 p q$
$= (9 p + 5 q)^{2}$

(iii)
${(\frac{4}{3} m - \frac{3}{4} n)}^{2} + 2 m n$

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

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

$= \frac{16}{9} m^{2} + \frac{9}{16} n^{2}$

(iv)
$(4 p q + 3 q)^{2} - (4 p q - 3 q)^{2}$
$= (4 p q)^{2} + (3 q)^{2} + 2 (4 p q) (3 q) - ((4 p q)^{2} + (3 q)^{2} - 2 (4 p q) (3 q))$
$= (4 p q)^{2} + (3 q)^{2} + 2 (4 p q) (3 q) - (4 p q)^{2} - (3 q)^{2} + 2 (4 p q) (3 q)$
$= 24 p q^{2} + 24 p q^{2}$
$= 48 p q^{2}$

(v)
$(a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a)$
$= a^{2} - b^{2} + b^{2} - c^{2} + c^{2} - a^{2}$
$= 0$

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Similar questions

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

Q. Show that:
(i) (3x + 7)² − 84x = (3x − 7)²
(ii) (9a − 5b)² + 180ab = (9a + 5b)²
(iii)

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

(iv) (4pq + 3q)² − (4pq − 3q)² = 48pq²
(v) (a − b)(a + b) + (b − c)(b + c) + (c − a)( c + a) = 0

Q. Show that :

(i) {(3 x + 7)}^{2} - 84 x = {(3 x - 7)}^{2}

(i i) {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 48 p q^{2}

(i i i) (a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0

Show that :

$(i) {(3 x + 7)}^{2} - 84 x = {(3 x - 7)}^{2}$

$(i i) {(4 p q + 3 q)}^{2} - {(4 p q - 3 q)}^{2} = 48 p q^{2}$

$(i i i) (a - b) (a + b) + (b - c) (b + c) + (c - a) (c + a) = 0$ [3 MARKS]

Q. Show that:

(9 p - 5 q)^{2} + 180 p q = (9 p + 5 q)^{2}

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Show that:(i) (3x+7)2−84x=(3x−7)2(ii) (9p−5q)2+180pq=(9p+5q)2(iii) (43m−34n)2+2mn=169m2+916n2(iv) (4pq+3q)2−(4pq−3q)2=48pq2(v) (a−b)(a+b)+(b−c)(b+c)+(c−a)(c+a)=0