Prove that $8 {(2 a + 3 b)}^{3} - 12 {(2 a + 3 b)}^{2} i s$ $4 {(2 a + 3 b)}^{2} (4 a + 6 b - 3)$

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Solution

Just take the common terms aside and simplify;
$8 (2 a + 3 b)^{3} - 12 (2 a + 3 b)^{2}$
$4 (2 a + 3 b)^{2} [2 (2 a + 3 b) - 3]$
$4 (2 a + 3 b)^{2} [4 a + 6 b - 3]$

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State whether True or False

Factorization of

8 (2 a + 3 b)^{3} - 12 (2 a + 3 b)^{2}

4 (2 a + 3 b)^{2} (4 a + 6 b - 3)

Factorise :

$(i) 6 x^{3} - 8 x^{2} (i i) 35 a^{3} b^{2} c + 42 a b^{2} c^{2} (i i i) 36 x^{2} y^{2} - 30 x^{3} y^{3} + 48 x^{3} y^{2} (i v) 8 (2 a + 3 b)^{3} - 12 (2 a + 3 b)^{2} (v) 9 a (x - 2 y)^{4} - 12 a (x - 2 y)^{3}$

a, b, c, d

are in continued proportion, prove that

2 a + 3 d : 3 a - 4 d = 2 a^{3} + 3 b^{3} : 3 a^{3} - 4 b^{3}

Q. Simplify:

(3a + 4)(2a − 3) + (5a − 4)(a + 2)

Q. Factorize:

6 (2 a + 3 b)^{2} - 8 (2 a + 3 b)

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Prove that 8(2a+3b)3−12(2a+3b)2is 4(2a+3b)2(4a+6b−3)

Just take the common terms aside and simplify; 8(2a+3b)3−12(2a+3b)2 4(2a+3b)2[2(2a+3b)−3] 4(2a+3b)2[4a+6b−3]

Prove that $8 {(2 a + 3 b)}^{3} - 12 {(2 a + 3 b)}^{2} i s$ $4 {(2 a + 3 b)}^{2} (4 a + 6 b - 3)$

Just take the common terms aside and simplify;
$8 (2 a + 3 b)^{3} - 12 (2 a + 3 b)^{2}$
$4 (2 a + 3 b)^{2} [2 (2 a + 3 b) - 3]$
$4 (2 a + 3 b)^{2} [4 a + 6 b - 3]$