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vodka [1.7K]
3 years ago
12

-2(70-4exponet3)+10 What do i do first

Mathematics
1 answer:
aleksandrvk [35]3 years ago
3 0
HERE IS THE ORDER
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Read 2 more answers
7. Verify the following: i) (ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0 ii) (a + b + c) (a² + b² + c² –
blsea [12.9K]

<u>1</u><u>.</u><u> </u><u>(ab + bc) (ab – bc) + (bc + ca) (bc – ca) + (ca + ab) (ca – ab) = 0</u>

We know that (a+b)(a-b) = a²-b²

(ab + bc)(ab -bc) can be written as a²b² - b²c²

(bc + ca)(bc -ca) can be written as b²c² - c²a²

(ca + ab)(ca - ab) can be written as c²a² - a²b²

→ a²b² - b²c² + b²c² - c²a² + c²a² - a²b²

→ a²b² - a²b² - b²c² + b²c² - c²a² + c²a²

→ 0

<u>2</u><u>.</u><u> </u><u>(a + b + c) (a² + b² + c² – ab – bc – ca) = a³ + b³+ c³ – 3abc</u>

→ a³ + ab² + ac² -a²b - abc -ca² + a²b + b³ + bc² - ab² - b²c - abc + a²c + b²c + c³ - abc - bc² - c²a

→ a³ + b³+ c³ + (- abc - abc - abc) + (ab² - ab² )+ (ac² - ca² ) -(a²b + a²b )+ (bc² - bc² )+ (a²c - c²a) + (b²c - b²c)

→ a³+b³+c³ - 3 abc .

<u>3</u><u>.</u><u> </u><u>(p – q) (p² + pq + q²) = p³ – q³. </u>

→ p³ + p²q + pq² - p²q - pq² - q³

→ p³ - q³ +(p²q - p²q) + (pq² - pq²)

→ p³ - q³

7 0
4 years ago
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