Question

what must be subtracted From 3a²-6ab-3b²-1 to get 4a²-7ab-4b²+1

Answer 1

Hope it helps you............

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Answer 2

$Let \: P \: should \:be \: subtracted \\from \:3a^{2}-6ab-3b^{2} - 1 \:to \:get \\4a^{2} - 7ab - 4b^{2} + 1$

$\implies (3a^{2}-6ab-3b^{2} - 1 )- P = 4a^{2} - 7ab - 4b^{2} + 1$

$\implies (3a^{2}-6ab-3b^{2} - 1 )-( 4a^{2} - 7ab - 4b^{2} + 1) = P$

$\implies 3a^{2}-6ab-3b^{2} - 1 -4a^{2} + 7ab + 4b^{2} - 1 = P$

$\implies 3a^{2} - 4a^{2} -6ab + 7ab - 3b^{2} + 4b^{2} -1-1 = P$

$\implies -a^{2} + ab + b^{2} -2 = P$

Therefore.,

$\green { -a^{2} + ab + b^{2}-2\: should}\\\green{ be \: subtracted }\:from \\3a^{2}-6ab-3b^{2} - 1 \:to \\get \: 4a^{2} - 7ab - 4b^{2} + 1$

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