Math, asked by pinkie3, 1 year ago

Plz answer this question

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Answered by GayatriPawar
1
v. 4a^2 + 9b^2 + 16c^2 + 12ab - 24bc - 16ca
=(2a)^2 + (3b)^2 + (-4c)^2 + [2 × 2a × 3b] +
[2 × 3b × (-4c)] + [2 × (-4c) × 2a]
IDENTITY:
(x+y+z)^2= x^2 + y^2 + z^2
+2xy + 2yz + 2zx

(2a + 3b -4c)






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pinkie3: thansk
pinkie3: I mean thanks
GayatriPawar: your welcome
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