On factorising (3a^2bc+9ab^2c+21abc^2)
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2
Step-by-step explanation:
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Answered by
1
Answer:
Answer
30(a
2
bc+ab
2
c+abc
2
)
=2×3×5[(a×a×b×c)+(a×b×b×c)+(a×b×c×c)]
=2×3×5×a×b×c(a+b+c)
Thus 30(a
2
bc+ab
2
c+abc
2
)÷6abc
=
2×3×abc
2×3×5×abc(a+b+c)
=5(a+b+c)
Alternatively 30(a
2
bc+ab
2
c+abc
2
)÷6abc
=
6abc
30a
2
bc
+
6abc
30ab
2
c
+
6abc
30abc
2
=5a+5b+5c
=5(a+b+c)
Step-by-step explanation:
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